Friction Friction force exists when there is relative motion between two contact surface. The friction force is self adjusting force, it can adjust its magnitude to any value between zero and limiting (maximum) value i.e. 0 ≤ f ≤ fmax There are three types of frictional forces - Static frictional force. - Kinetic frictional force. - Rolling frictional force. Static Frictional Force The frictional force between the two surface before the relative motion actually starts is known as static friction. Laws of Static Friction (i) The force of friction is proportional to the normal force that keeps the two surfaces in contact with each other. The static frictional force is given by fs ≤ μ sN Where μ s is static coefficient of friction. The less than equal to sign in the above equation represents adjusting nature of the force of static friction. The equality sign in the equation holds only when it has its maximum value. (ii) It is independent of area of contact between two surfaces. Kinetic Frictional Force The frictional force when the surface in contact are in relative motion is known as kinetic friction or dynamic friction. Laws of Kinetic Friction (i) The force of kinetic friction is proportional to the normal force, which presses the two surfaces together. fk = μkN Where μk is the coefficient of kinetic friction. (ii) It is almost independent of surface area of contact. (iii) It is independent of speed of sliding. Angle of Friction The angle which the resultant of limiting friction fs and normal reaction R makes with normal is known as angle of friction and denoted by λ. tan λ = fs/R = μ Angle of Repose This angle is relevant to inclined plane. If the body is placed on an inclined plane and it is just on the point of sliding down, then the angle of inclination of the plane with horizontal is called angle of repose (α). From fig. f = Mg sinα ----(1) and R = Mg cosα ------(2) Applying (1) /(2) f/R = tan α = μ .: α = λ i.e. Angle of repose = angle of limiting friction. Motion on a Rough Horizontal Surface (i) If F is the minimum force required in order to just slide a body, then F = μsR = μs mg where μs is coefficient of static friction. (ii) In order to maintain a body sliding with uniform speed, the force required is F = μkR = μk mg where μk is coefficient of kinetic friction. Motion Down the Plane The resultant force with which a body slides down an inclined plane is P = mg sinθ - f = mg sinθ - μkN or ma = mg sin θ - μk mg cosθ (... N = mg cosθ) Hence, acceleration down the plane in such case a = g(sinθ - μkm cosθ) Q. A block of mass 100 kg is kept on a rough inclined surface having an angle of inclination 30o from horizontal. If body starts from rest, then calculate downward acceleration (coefficient of friction 0.2) Soln: We know that downward acceleration a = gsinθ - μkmgcosθ = 9.8 × 1/2 - 0.2 × 9.8 × √3/2 = 3.2 m/s2 Motion Up the Plane (i) The minimum force required to move a body up the plane P = mg sinθ + f = mg sinθ + μkR = mg sin θ + μkR cosθ (ii) Minimum force required to prevent the body from moving down the plane P = mg sin θ - f = mg (sin θ + μs cosθ) Q. A box is placed on an inclined plane and has to be pushed down. The angle of inclination to angle of friction is (a) equal to angle of friction. (b) more than angle of friction. (c) equal to angle of repose. (d) less then angle of repose. Ans: (d) Hint: When the angle of inclination is equal to angle of repose, the body just slides down the plane. But when the angle of inclination is greater than angle of repose, the body begins to accelerate down the plane. Q. To avoid slipping while walking on ice, one should take smaller steps because of (a) friction of ice is large. (b) large normal reaction. (c) friction of ice is small. (d) small normal reaction. Ans: (c) Note: Due to small friction of ice, we can not obtain larger reaction. So, to avoid slipping on ice we should take smaller steps to produce larger normal reaction. Note: It is misconception that the frictional force is always given by F = μN. But friction force is a self adjusting force. It can adjust its magnitude anywhere in between 0 and fmax , i.e. 0 ≤ f ≤ fmax - If net tangential force F tending to accelerate the body is less than or equal to the maximum friction force, then the magnitude of friction force is equal to net force i.e. if F ≤ fmax then f = F and f ≠ μN Since F= 20 N is less then fmax = μN = 0.4*100 = 40 N Therefore, f = 20 N - If net tangential force F accelerating the body is more than maximum friction force, then the magnitude of friction force is equal to the maximum friction force. ie. F > fmax , f = fmax = μN Since F= 60 N is more then fmax = μN = 0.4*100 = 40N. Therefore, f = fmax = 40 N Q. A scooter is moving on a straight horizontal surface with a velocity u, calculate the shortest distance in which the scooter can be stopped, if coefficient of friction between tires and roadis μ. Soln: Friction force, f = μR = μ mg or ma = μ mg .: retardation, a = μg We know, v2 = u2 - 2as .: s = v2/2μg Q. A horizontal force of 12 N pushes a block weighing 5N against a vertical wall. The coefficient of static friction between the block and the wall is 0.6 and coefficient of kinetic friction is 0.4. Assuming the block is not moving initially, (i) will the block is start moving? (ii) what is the force exerted on the block by the wall? (a)12 N (b)5 N (c)7.2 N (d)13 N Soln: (i) the block will not movie. (ii)Ans: (d) force applied = 12 N .: fmax = μR = 0.6 × 12 = 7.2 N This shows that fmax > W So, frictional force (f) = 5N Net contact force F = √(R2 + f2) = √(122 + 52) = 13 N Q. A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. The coefficient of friction between the two blocks is μ1 and that between the block of mass M and horizontal surface is μ2. What maximum horizontal surface can be applied to the lower block so that the two blocks move without separation? (a)(M + m)(μ2 -1)g (b)(M - m)(μ2 - μ1) g (c)(M - m)(μ2 + μ1)g (d)(M + m)(μ2 - μ1)g Ans: (d) Here, the force applied should be such that the force on block of mass m should not exceed the force of friction (μ1mg). Let, the system move with acceleration a, then F - μ2 (M + m)g = (M + m) a -----(1) for block of mass m, F1 = ma = μ1mg => a = μ1 g --------(2) From (1) & (2) we get F - μ2 (M + m) g + (M + m) (μ1 g) = (M + m) g (μ2 + μ1) .: F = (M + m) (μ1 + μ2) g Note: Maximum force applied to lowest block so that all the blocks move without separation is F = (∑m) (∑μ) g .: F = (m1 + m2 + m3 + m4) (μ 1 + μ 2 + μ 3 + μ 4) g Q. Two blocks of mass 6 kg and 4 kg are kept as shown in fig on a smooth horizontal surface. When maximum force 20 N is applied to lowest block, then they move together. Calculate coefficient of friction between the surfaces of two blocks. (a)0.1 (b)0.2 (c)0.3 (d)0.4 Ans: (b) Soln: F = (∑m) (∑μ) g or F = (m1 + m2) (μ1 + μ2) g or 20 = (4 + 6) (μ + 0) 10 .: μ = 20/100 = 0.2 (.: for smooth surface μ2 = 0) Q. Determine the magnitude of fricitonal force in each of the following cases (i) (ii) (iii) Soln: (i) W = mg = 5 * 10 = 50 N Frictional force (max) fmax = μR = 0.2 * 100 = 20 N Here, fmax < W, therefore frictional force = 20N (ii) W = mg = 5 * 10 = 50 N Frictional force (max) fmax = μR = 0.2 * 500 = 100 N Here, fmax > W, therefore frictional force = 50N (iii) W = mg- 100 sin30 = 5 * 10 - 100*1/2 = 0 N Frictional force (max) fmax = μR = 0.2 * 100 cos30 = 10 √3 N Here, tangential downward force(i.e. weight) which tends to accelerate the boby is zero, so no frictional force exits. so frictional force, f = 0 Work, Energy & Power Work Under constant force : W = F . S = F S cosθ = Ei - Ef = ΔE (change in energy) Under variable force : W = ʃ F . ds = ΔE >>Work done is scalar quantity. So it may be positive or zero. >>Whenever work done is +ve, work is done by the body and its energy gets decreased. >>Whenever work done is -ve, work is done on the body and its energy gets increased. Q. A body of mass m is raised from earth surface to a height equal to radius of earth. What is the work done on the body, if acceleration due to gravity on earth surface is g. (a) - mgR (b) mgR (c) mgR/2 (d) mgR/4 Ans: (c) Soln dw = F dx = GmM/x2 dx :. W= ʃR2R GmM/x2 dx = GMm [-1/x ]R2R = GMm (-1/2R + 1/R) = GMm/R Short Hint: ΔP = mgh (R/R + h) = mgR (R/R + R) = mgR/2 Q. A ball falls from height 10 m. On rebounding , it losses 30% energy. The ball goes upto the height of (a) .5 m (b) .7 m (c) .6 m (d) .8m Ans: (b) Soln: final energy = 70/100 × initial energy mgh2 = 7/10 × mgh1 h2 = 0.7 m Q. A body is displaced from position (i + 3j + k)m to position (2i + j + k)m under the action of constant force F = (i + j + k)N. The work done will be (a) 1 J (b) 2 J (c) -1 J (d) 3 J What work will be done in moving a body of mass m upwards along an inclined plane of angle θ through a distance s. The frictional force is Fx The function of F is to overcome mg sin θ and FK or, F = mg sin θ + FK :. W = (mg sin θ + FK)s Cases: >>If θ = 0o, ω = FK S i.e. work is said to be done against the friction. >>If θ = 90o, Then FK = 0, S = h or W = mg × 1 × h or W = mgh i.e. work is said to be done against the gravity. >>If body moves downward F + mg sin θ + FK :. F = ( FK - mg sin θ ) W = (FK - mg sin θ)s >>If body moved downward vertically, the work done on the body W = -mgh In above question, if the force is not constant but varies from zero to F uniformly. The work done is given by Fav = (0 + F)/2 = F/2 W = Fav . S = FS/2 Q. A pole of mass m and length l is lying on the ground. What work must be done in erecting the pole vertically on the ground. Soln W = mgh = mgl/2 Q. A chain of mass m and length l lying on a table. If 1/nth part of the total length is hanging over the edge. What work must be done in pulling up whole of the chain? Soln mass per unit length = m/l mass of l/n length = m/l × l/n = m/n :. mass to be pulled = m/n effective height, h = (l/n)/2 = l/2n . :. W = mgh = m/n × g × l/2n = mgl/2n2 :. W = mgl/2n2 Q. A uniform chain of length L and mass M lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is: (a) mgL (b) mgL/3 (c) mgL/9 (d) mgL/18 Ans: (d) :. W = mgl/2n2 = mgl/2 × 32 Power Rate of doing work = (work done)/(time taken) Avg. Power = It is the ratio of total work done to the total time taken. Pav = W/t = F.S/t Instaneous Power : It is the ratio of infinitely small work done at an instant to the corresponding time taken. Pinst = Δ W/Δ t = F. Δ S/Δ t = F. v :. Pinst = F v When velocity is constant, Pinst = Pav. Q. A constant force of 20 N acts on a body of mass 10 kg. What would be the average power in 10 sec. What will be the power at the instant of 10 sec. Soln: 1st case : a = E/m = 20/10 = 2 m/s2 S = 1/2 at2 = 1/2 × 2 × 102 = 100 m Pav = P.S/t = (20 × 100)/10 = 200 Watt. 2nd case : Vel. at the end of 10 sec = V = u + 2 × 10 = 20 m/s. Pinst = F v = 20 × 20 = 400 Watt. Q. What power is required to move a body of mass m upward along an inclined plane of an angle θ with constant velocity v. The frictional force is EK. Soln: => W = (mg sin θ + EK) S :. P = (mg sin θ + EK)V Q. What power must be developed by the engine of car of mass 1000 kg to move with vel. 10 m/s along the inclined plane having angle of inclination sim-1 1/5. If the frictional force developed between can and the inclined surface is 5% of the weight so the car. (a) 20 KW (b) 25 KW (c) 30 KW (d) 10 KW Ans: (b) Soln: => P = (1000 × 102 × 1/5 + 1000 × 10 × 5/100) 10 = (2000 + 500) × 10 = 2500 × 10 = 25 KW K.E. of moving object KE = 1/2 mv2 = 1/2 m2v2/m = P2/2m :. EK = P2/2m P.E. :- A mass in the field of gravity possesses potential energy due to force of gravity. The P.E. of a body a body depends upon the distance from the centre of the earth. In the fig, work done in taking the mass from level P to level Q will be: W = mgh or, PEQ - PEP = mgh This work done stores in the body in the form of gain in potential energy. If we take PE of P level as zero. then, PEQ = mgh. Energy conservation states that the sum of PE and KE of a body remains constant. If a body is projected up. the loss of KE changes to gain in PE. similarly, if a body is fallen from height, the loss of PE is changed into gain in KE. Circular Motion Q. Uniform circular motion is motion with (a) constant velocity (b) constant acceleration (c) constant speed (d) both (b) and (c) Ans: (c) Circular Motion Motion in circular path with constant or variable speed is known as circular motion. Uniform Circular Motion Motion in circular path with constant speed. - It is variably accelerated motion. - It is a periodic motion in which acceleration is directed towards the center of circular path. - Speed, KE, angular velocity, time period, angular momentum remain unchanged. - Acceleration, force, velocity, and momentum each changes in direction continuously but their magnitude remains constant. - Velocity and momentum act tangentially while acceleration and force act racially inward towards centre. - Velocity or momentum is always perpendicular to acceleration or force . - Angular acceleration, moment of force about axis through centre i.e. torque and work done are zero. Variable Circular Motion Motion in circular path with variable speed. - In variable circular Motion, velocity changes in magnitude as well as in direction. Hence acceleration is not directed towards centre racially. Note: Angular displacement, θ = arc/radius, unit radian is axial vector when small and scalar when large. If n in the no. of revolutions, then angular displacement is equal to 2πn (Angular displacement θ = 2 πn) Q. A particle makes 1.5 revolutions in a circular path of radius 2 cm. Find the angular displacement. Hint: θ = 2πn = 2 π × 1.5 = 3π Angular Velocity ω = Δθ/Δt = unit rad/s It is the axial vector. Tachometer measures angular speed. Q. A particle is moving along a circular path of radius 2 m and with uniform speed of 5 m/s. What will be the change in velocity when the particle completes (i) half of the revolution? (a) 0 (b) 10 m/s (c) 10√2 m/s (d) 10/√2 m/s Ans: (b) Soln: Δv = 2v sin θ/2 =2 × 5 × sin 90o = 10 m/s (ii) 1/4 th of revolution? (a) 5√2 m/s (b) 0 m/s (c) 10 m/s (d) 10/√2 m/s Ans: (a) Soln: Δv = 2 × 5 × sin 45o = 5√2 m/s Q. A car is moving with speed 30 m/s on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m/s2. The acceleration of the car is (a) 2 m/s2 (b) 2.7 m/s2 (c) 1.8 m/s2 (d) 9.8 m/s2 Ans: (b) Hint: a = √(at2 + ac2) where ac = V2/R & at = 2 m/s2 Motion of Cyclist N cos θ = mg -----------------(1) N sin θ = mv2/R ------------------(2) From (1) and (2) we get, tan θ = v2/Rg If μ is the coeffient of friction, then for no skidding of cycle or overturning of cyclist μ ≥ tan θ = v2/Rg or, v ≤ √(μRg) So, maximum speed for no side slip, Vmax = √(μRg) Q. A man is cycling on a circular track of radius 20 m. If the frictional coeffient between the road and the tyre is 0.5, at what maximum speed should the man cycle not to slip to the side. (a) 10 (b) 20 (c) 30 (d) 40 Ans: (a) Hint: Vmax = √(μRg) Motion in Vertical Circle This is the example of variable circular motion. Velocity changes in magnitude and direction. A particle of mass m is whirled in a circle of radius r, then tension in string T - mg cos θ = mv2/R => T = mv2/R + mg cos θ At lowest point A, θ=0o, v = vA, T = mvA2 /R + mg = Tmax At the highest point i.e. B, θ = 180o, v = vB T = mvB2 /R - mg = Tmin At point D, θ = 90o T = mv2 /R Different between maximum and minimum tension Tmax - Tmin = m(vA2 - vB2) /R + 2mg = (m/R) × 4Rg + 2mg Tmax - Tmin = 6 mg Condition to make body in vertical circle TB ≥ 0 or, mvB2 /R - mg ≥ 0 => vB ≥ √Rg TB = Tmax= Tmin + 6mg or, TB = 0 + 6 mg (TB = 6 mg) vB = √Rg is known as critical velocity at the highest point. Applying law of conservation of energy EA = EB 1/2 mvA2 + 0 = 1/2 mvB2 + mg 2R 1/2 vA2 = 1/2 Rg + 2Rg => vA = √5Rg Condition for Oscillation in a Vertical Circle The pendulum will oscillate if its velocity at lowest point L is such that the velocity reduces to zero at M1 or M2 . Applying law of comesvation of energy 1/2 mv2 = mgR => v = √2Rg i.e. velocity at lowest point L should be equal to or less then √2Rg So, the particle leaves the circular path iff √2Rg < vA < √5Rg Q. A body tied to a light inextensible string 10/3 m long is whirled in vertical circle such that minimum tension in string is 4 times its maximum tension. The velocity of body at the highest point will be (a) √(100/3) m/s (b) √50 m/s (c) 10 m/s (d) 10/3 m/s Ans: (c) Hint: Tmax - Tmin = 6 mg or, 4 Tmin - Tmin = 6 mg or, Tmin = 2 mg or, mv2 /r - mg = 2 mg (we know Tmin = mv2 /r - mg) or, mv2 /r = 3mg or, v = √(3rg) = √(3lg) = √(3 × 10/3 × 10) = 10 m/s Q. A bucket filled with liquid is whirled in vertical circle of radius R with the help of a rope. In order to prevent falling of liquid out of the bucket the minimum velocity at highest point will be (a) √3gR m/s (b) √gR m/s (c) √2gR m/s (d) √5gR m/s Ans: (b) Q. A vehicle of mass m moving with constant speed v along concave road of radius of curvature R is at lowest point. The reaction on the vehicle by the road will be (a) mv2 /R - mg (b) mg - mv2 /R (c) mv2 /R + mg (d) mg Ans: (c) Hint: N - mg = mv2 /R => N = mv2 /R + mg Q. A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point such that (a) h = R (b) h = R/3 (c) h = R/2 (d) h = 2R Ans: (b) Hint: At point P, N = 0 mg cos θ = mv2 / R or, mg × (R - h)/R = mv2 / R or, g (R - h) = 2gh or, R - h = 2h or, 3h = R . : h = R/3 Q. In above question, at what angle θ with vertical, the block just slides tangentially loosing the contact of that hemisphere. (a) cos-1(1/3) (b) cos-1(2/3) (c) tan-1(2/3) (d) tan-1(1/3) Ans: (b) cos θ = (R - h)/R = (3h - h)/3h = 2h/3h = 2/3 θ = cos-1(2/3) Q. From what height h, a body should be released along a frictionless plane so that it will just revolve in vertical circle of radius R after falling the ground (a) 2R (b) 5/2 R (c) 3/2 R (d) 3R Ans: (b) Soln: v = √2gh, but vA = √5Rg so, vA = v => 5Rg = 2gh =>hmin = 5/2 R Gravitation Gravitation: Phenomenon of mutual attraction between any two bodies due to mass. Gravitational force: - exists due to mass. - is always attractive. - is mutual. - acts along line joining their centre. - depends on masses of the bodies. - depends on separation between their centers. - does not depends on medium between these bodies. - It is the central force hence conservative, work done by it is independent of path followed. F = Gm1m2 /r2 G = 6.67 × 10-11 Nm2 /kg 2 Q. The Gravitational force between two steel balls of radius R each touching their surfaces is F. What will be the Gravitational force between two steel balls of radius 3R each touching their surfaces. (a) F/9 (b) F/81 (c)9 F (d) 81F Ans: (d) F = Gm1m2 /R2 = Gm2/R2 => F α m2/R2 => F α V2/R2 => F α R6/R2 .: F α R4 F = (3R)4 F = 81 R4 = 81F Note: If two spheres of same mass, radius, material are put in contact, the gravitational attraction between them is directly proportional to 4th power their radius. Q. A lump of mud is divided into two lumps of masses m1 and m2 such that the gravitational attraction between these lumps will be maximum, then (a) m1 = m2 (b) m1 = 2m2 (c) m1 = 4m2 (d) m1 and m2 may be in any ratio Ans: (a) F = [Gm(M - m)]/R2 = (G/R2) (mM - m2) or, dF/dm = G/R2 (M - 2m) or, d2F/dm2 = - G/R2 × 2 => -ve => F = Fmax For maxima, dF/dm = 0 i.e. G/R2 (M - 2m) = 0 => M = 2m => m = M/2 => m1 = m2 Gravitation Field Space around a body where it exerts gravitational force on the body. Gravitational Field Intensity E = F/m = GM/ R2 = dV/dr where V is gravitational potential. Gravitation Potential V = W/m = - ʃEdr (i) For a point man M at distance r from it V = -GM/r (ii) For a solid sphere of uniform density - outside the sphere i.e. r > R V = -GM/r - on the surface i.e. r = R V = -GM/R = -gR - Inside sphere i.e. r < R V = -GM/2R3 (3R2 - r2) - at the centre i.e. r = 0 V = -3GM/2R = (-3/2) gR Gravitational potential inside gravitational field is always -ve => force is attractive. Gravitational potential increases as we move away from earth surface. (iii) For hollow sphere Gravitational potential - outside the sphere V = GM/r - on the surface of sphere V = -GM/R - Inside the sphere V =- GM/R (everywhere constant) Gravitational field intensity - outside the sphere E = GM/ r2 - on the surface of sphere E = GM/ R2(max) - Inside the sphere E = 0 (everywhere) Q. The gravitational potential at the center of earth surface will be (a) - gR (b) -3/2 gR (c) 0 (d) ∞ Ans: (b) Q. The gravitational potential at any point at distance r from the center of the body is V = -k/rn. The gravitational field strength at that point will be (a) -k/rn (b) -nk/rn (c) -nk/rn+1 (d) (n +1)k/rn+1 Ans: (c) E = -dV/dr or, E= d/dr (k/rn) = k d/dr r-n = k(-n) r-n+1 = -kn 1/ rn+1 Gravitational Potential Energy (U) U = mgh / (1+ h/R) If h n=5 for g = g/5 Therefore, x = R - R/5 = 4/5 R Q. The acceleration due to gravity at 20 km height from earth surface is g. What will be the acceleration due to gravity at depth 40 km from the surface of the earth? (a) g (b) g/2 (c) 2g (d) g/3 Ans : (a) Hint : x = 2h for same g Q. What is the ratio of acceleration due to gravity at height R/2 and at depth R/2? (a) 1:1 (b) 4:9 (c) 9:8 (d) 8:9 Ans : (d) Soln: We know At height h from earth surface gh = g (1 + h/R)-2 ------(i) And at depth x from earth surface gx = g (1 - x/R) --------(ii) Dividing (i) by (ii) gh / gx = (1 + h/R)-2 / (1 - x/R) = (1 + 1/2)-2 / (1 - 1/2) = (3/2)-2 / (1/2) = 2 /(9/4) = 8/9 (3) Variation of g on the earth surface On the surface of the earth, value of g changes due to shape of earth and rotation of earth. Variation on surface due to shape g = GM/R2 α 1/R2 RE > RP (Re = RP + 21 km) .: gP > gE gP - gE = GM(1/RP2 - 1/Re2) = 0.018 m/s2 = 1.8 cm/s2 Variation due to rotation (effect of latitude/rotation) g = g - ω2R cos2θ ω = 2π / (24 × 3600) = 2π /86400 rad/s At equator : θ = 0o ge = g - ω2R ; ge =9.780 m/s2 At pole : θ = 90o gP = g (no rotational effect) ; gP = 9.832 m/s2 Δg = gP - ge = 0.034 m/s2 = 3.4 cm/s2 If the earth stops rotating, then ω = 0 g = g everywhere. So, the value of g increases in equator by Rω2 and g is unaffected at poles. Q. Determine the speed with which the earth rotates so that a person on equator would weight 2/5 as much as at present. Soln: ge = g - ω2 or, 2/5 g = g - Rω2 . : ω = √(3/5 g/R) Q. What would be the angular speed of rotation of earth so that particle at earth equator will be weightless. (a) 7.29 × 10-5 rad/s (b) 1.25 × 10-5 rad/s (c) 7.29 × 10-3 rad/s (d) 1.25 × 10-3 rad/s Ans : (d) Soln: At equator g = g - Rω2 for weightless, g = 0 g = Rω2 ω = √(g/R) = 1/800 rad/s = 1.25 × 10-3 = 17 times present angular velocity. Q. If the radius of earth decreases by 50% on shrinking. The value acceleration due to gravity on its surface increases by (a) 50% (b) 100% (c) 200% (d) 300% Ans : (d) Soln: g = GM/R2 , On shrinking g = GM/(R/2)2 = 4 g % increases (g - g)/g × 100% = 300% Q. Two planets have equal densities and their radii are in the ratio 1:2. The ratio of acceleration due to gravity on their surface will be (a) 1:2 (b) 1:4 (c) 1: √2 (d) 4:1 Ans : (a) Hint: g = GM/R2 = 4/3 πGρR . : g α R Satellite and Escape Velocity Orbital Velocity of a Satellite Vo = √(GM/r) = √(gR2/r) = √[(4πGρR3)/3r] where r = R + h Orbital velocity depends upon (i) radius of orbit. (ii) height of the orbit above the surface of planet. (iii) Mass, radius, density, and acceleration due to gravity of the planet. - Orbital velocity of a satellite does not depend on mass, shape, size, radius of satellite. - Satellite revolving in orbit close to the earth surface is fastest satellite. r ≈ R Vo = √(GM/R) = √(gR2/R) = √(gR) = 8 km/s This velocity is also called its cosmic velocity of earth. Q. If ρ be the mean density of earth, then time period of revolution of a satellite orbit close to the earth will be (a) 2π √(ρ/G) (b) √(3π /Gρ) (c) 2π √(3/Gρ) (d) √(π /3Gρ) Ans : (b) For a Satellite revolving in an orbit of radius r (i) Orbital speed, Vo = √(GM/r) = √(gR2/r) where r = R + h (ii) Angular velocity, ωo = vo/r = √(GM/r3) = √(gR2/r3) (iii) Time period, T = 2π/ωo = 2π√( r3/GM) = 2π√( r3/ gR2) (iv) Frequency of revolution, f = (1/2π)√( GM/r3) = (1/2π)√(gR2/r3) (v) Kinetic energy, KE = 1/2 mvo2 = mGM/2r = GMm/2r = gR2m/2r (vi) Potential energy, U = -GMm/r = - gR2m/r (vii) Total mechanical energy = PE + KE = -GMm/2r = -gR2m/2r (-ve sign shows the binding energy of satellite.) For a satellite close to earth r ≈ R Vo = √(gR) = 8 km/s ωo = √(g/R) = 1.25 × 10-3 rad/s T = 2π√(R/g) = 83.7 ≈ 84 minutes Q. The KE of a satellite revolving in an orbit is Eo. Its total energy in that orbit will be (a) Eo (b) 2Eo (c) -Eo (d) -2Eo Ans : (c) Hint: EK = GMm/2r = Eo U = -GMm/r = -2Eo ET = GMm/2r = -Eo Q. If a body is released from a satellite revolving in a stationary orbit around the earth, then the body (a) flies away from earths surface gravitational field. (b) Moves straight tangentially. (c) Starts revolving in some orbit with satellite. (d) Falls to earth along spiral path. Ans : (c) Q. A satellite in an orbit is revolving with KE Eo. What minimum energy is to be applied to it so that it just escapes the earth gravitational field? (a)2Eo (b)√2Eo (c)Eo/2 (d)Eo Ans : (d) Soln: For orbital velocity KE = 1/2 Vo2 = 1/2 m Rg = Eo [:.Vo = √(Rg)] For escape velocity KE = 1/2 Ve2 = 1/2 m (2Rg) = 2Eo [:.Vo = √(2Rg)] Eapplied = 2Eo - Eo = Eo Note:The linear speed of any particle on earth surface at latitude θ is ωR cosθ where ω = 2π /(24 × 3600) rad/s and R = 6.4 × 106 m The linear velocity of particle on earths equator will be v = ωR Geostationary /Communication / Synchronous Satellites : Satellites revolving around earth in gravitational field in equatorial plane for which time of revolution is equal to the time period of revolution of earth. Escape Velocity The minimum velocity with which if a body is projected, escapes out of the gravitational field. Ve = √(2gR) [1/2 mv2 = GMm/R] The escape velocity does not depend upon direction of projection and mass of the projected body. Q. The escape velocity of the projectile on earths surface is 11.2 km/s. A body is projected out with thrice this speed. What is the speed of the body far from the earth? Ignore the presence of sun and other planet. Hint: 1/2 mve2 = -GMm/r + 0 ------(1) 1/2 mve2 = -GMm/r + 1/2 mv2 ------(2) Using (2) - (1) 1/2 m(ve2 - ve2) = 1/2 mv2 .: v = √(ve2 - v2) If body is thorn n times ve v = √(ve2 - ve2) = √( n2 - 1) ve v = √( n2 - 1) ve Points to Remember - Apparent wt. (Wa) is the reaction of the surface on which a body lies. - In freely falling system, time of simple pendulum is infinity. τ = 2π√(l/g) = 2π√(l/0) = ∞ - No energy is dissipated in keeping the satellite in orbit around a planet or work done to keep the satellite in orbit is zero. - The centripetal acceleration of satellite is equal to g. - Orbital vel. depends on mass of the planet and radius of orbit. - If the body is at height h above the surface of the earth, then the escape velocity Ve = √[2gh(R + h)] - Escape velocity, ve = √2 × orbital vel (vo) - If a body is orbiting around earth, then it will escape away, when its velocity increased by 41.8% or when K.E. is doubled. - If the radius of earth is increased by n times, keeping the density unchanged, the scope velocity will be n times the present value. - If the radius of earth increased by n times, keeping the mass constant, the escape velocity will be (1/√n) times the present value. - If a body freely falls from ∞ height, then it reaches the surface of the earth with velocity 11.2 km/s.
Posted on: Wed, 13 Aug 2014 17:03:04 +0000
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