In most of the Medical and Engineering Entrance test papers you will encounter at least a couple of questions on rotational motion. Let us consider the following questions:
(1) A solid cylinder initially at rest rolls down an inclined plane of angle θ and height ‘h’ without slipping. The linear velocity with which it will reach the bottom of the plane is
(a) √(3gh/4) (b) √(4gh/3) (c) √(4ghsinθ/3) (d) √(5gh/4) (e) √(2ghsinθ/3)
If ‘v’ and ‘ω’ are the linear and angular velocities respectively, ½Mv2 + ½Iω2 = Mgh where I is the moment of inertia and M is the mass of the cylinder. Substituting I = ½ MR2 and ω = v/R , we get v = √(4gh/3). Note that the angle θ mentioned in the problem is just a distraction.
(2) A simple pendulum bob of mass ‘m’ is drawn to one side so that the string is horizontal. It is then let free. When it crosses the mean position, the tension in the string is
(a) mg (b) 0.5mg (c) 1.5mg (d ) 3mg (e) 5mg
In problems of this type, you should note that the centripetal force (which is equal to mv2/r) acting on the body executing circular motion is the net force on the body. Therefore, centripetal force, mv2/r = T – mg, where r is the length of the pendulum and T is the tension in the string.Therefore, T= mv2/r + mg. But, ½ mv2 = mgr, on equating the initial gravitational potential energy of the bob to its kinetic energy in the mean position. From this, mv2/r = 2mg and hence T = 2mg + mg = 3mg.
Let us consider the following two questions involving the rotation of the earth:
(1) If the radius of the earth is changed to 1/√3 times the present value, the duration of the day (in hours) will be
(a) 72 (b) 41.6 (c) 24 (d) 12 (e) 8
This question is set to check your understanding of the law of conservation of angular momentum: I1ω1= I2ω2 where I1and I2 are the moments of inertia and ω1 and ω2 are the angular velocities of the earth before and after the contraction respectively. Substituting for I1 (= 2MR2/5) and I2 [= 2M (R2/3)/5] we obtain ω2 = 3ω1. Since the angular velocity changes to 3 times the initial value, the spin period of the earth (T= 2π/ω) changes two one-third of the initial value. So, the duration of the day will become 24/3 = 8 hours.
Questions of this type are often found in Medical and Engineering Entrance Test papers. Generally, if the radius of the earth becomes ‘n’ times the present value, the duration of the day becomes 24n2 hours. Remember this equation and write the answer in no time!
(2) If your weight while standing on the earth’s surface at the equator is to become zero, the earth should spin at nearly ------ times the present speed.
(a) 12 (b) 14 (c) 17 (d) 24 (e) 37
This is a simple question. If you are to become weightless due to the spin of the earth, the gravitational pull on you is to be balanced by the centrifugal force so that, mg = mRω2. From this ω=√(g/R) and the spin period T=2π/ω=2π√(R/g). On substituting for R = 6400 km and g = 9.8ms-2, the period works out to be 84.6 minutes. This is one-seventeenth the present period of 24 hours. So, the earth should spin at 17 times the present speed. [Option (c)].
(1) If the radius of the earth is changed to 1/√3 times the present value, the duration of the day (in hours) will be
(a) 72 (b) 41.6 (c) 24 (d) 12 (e) 8
This question is set to check your understanding of the law of conservation of angular momentum: I1ω1= I2ω2 where I1and I2 are the moments of inertia and ω1 and ω2 are the angular velocities of the earth before and after the contraction respectively. Substituting for I1 (= 2MR2/5) and I2 [= 2M (R2/3)/5] we obtain ω2 = 3ω1. Since the angular velocity changes to 3 times the initial value, the spin period of the earth (T= 2π/ω) changes two one-third of the initial value. So, the duration of the day will become 24/3 = 8 hours.
Questions of this type are often found in Medical and Engineering Entrance Test papers. Generally, if the radius of the earth becomes ‘n’ times the present value, the duration of the day becomes 24n2 hours. Remember this equation and write the answer in no time!
(2) If your weight while standing on the earth’s surface at the equator is to become zero, the earth should spin at nearly ------ times the present speed.
(a) 12 (b) 14 (c) 17 (d) 24 (e) 37
This is a simple question. If you are to become weightless due to the spin of the earth, the gravitational pull on you is to be balanced by the centrifugal force so that, mg = mRω2. From this ω=√(g/R) and the spin period T=2π/ω=2π√(R/g). On substituting for R = 6400 km and g = 9.8ms-2, the period works out to be 84.6 minutes. This is one-seventeenth the present period of 24 hours. So, the earth should spin at 17 times the present speed. [Option (c)].
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