PROBLEM: A 1.0 kg solid cylindrical pulley of radius 0.050 mhas a 0.30 m long ro

Question

PROBLEM: A 1.0 kg solid cylindrical pulley of radius 0.050 mhas a 0.30 m long rod (of negligible mass) attached to it with a0.50 kg mass (m1) at the end of the rod. The mass is initiallyvertically above the pulley, but is free to rotate about the centerof the pulley. The pulley is 0.30 m above a level frictionlesstrack where a 1.2 kg mass (m2) rests. When the rod rotatesdownward, m1 breaks away from the rod, and elastically collideshorizontally on the surface with m2, sending m2 along africtionless track; the track has a 0.20 m diameter vertical loopin it. Without additional input, gravity alone will not be able toswing m1 fast enough so that m2 can make it around the loop in thetrack, so another mass (m3) is attached by a rope that is wrappedaround the pulley and hangs so as to provide additionaltorque/energy input. (Note: I made mass m1 break away at the pointof collision so that we wouldn’t need to treat the pulley andm3 in the elastic collision.)
(i) What is the minimum mass m3 that can be used so that m2can navigate the loop in the track? (ii) What is the recoil velocity of m1?
I just can't seem to get this one. I'm not even sure where tobegin. PROBLEM: A 1.0 kg solid cylindrical pulley of radius 0.050 mhas a 0.30 m long rod (of negligible mass) attached to it with a0.50 kg mass (m1) at the end of the rod. The mass is initiallyvertically above the pulley, but is free to rotate about the centerof the pulley. The pulley is 0.30 m above a level frictionlesstrack where a 1.2 kg mass (m2) rests. When the rod rotatesdownward, m1 breaks away from the rod, and elastically collideshorizontally on the surface with m2, sending m2 along africtionless track; the track has a 0.20 m diameter vertical loopin it. Without additional input, gravity alone will not be able toswing m1 fast enough so that m2 can make it around the loop in thetrack, so another mass (m3) is attached by a rope that is wrappedaround the pulley and hangs so as to provide additionaltorque/energy input. (Note: I made mass m1 break away at the pointof collision so that we wouldn’t need to treat the pulley andm3 in the elastic collision.)
(i) What is the minimum mass m3 that can be used so that m2can navigate the loop in the track? (ii) What is the recoil velocity of m1?
I just can't seem to get this one. I'm not even sure where tobegin.

Explanation / Answer

(i)let the angular velocity of mass m2 be wtherefore we get w2 - wo2 =2 wo = 0 and = 2 radians = 2 * 3.14radians = 6.28 radians w2 - (0)2 = 2 or w = (2)1/2 the torque acting on mass m1 is = I * or w * l = I * or = (w * l/I) w = m1 * g,l = 0.30 m and I =(m1 * l2/3) or = (m1 * g * l/(m1 *l2/3)) = (3g/l) g = 9.8 m/s2 and l = 0.30 m the linear speed of mass m2 is v = r * w r = 0.050 m the collision between mass m1 and m2 iselastic therefore the momentum is conserved during elasticcollision (m1 + m3) * v = m2 * v or m3 = m2 - m1 = 1.2 - 1.0 =0.2 kg (ii)The recoil velocity of m1 is same as the recoilvelocity of m2.This is because the collision betweenm1 and m2 is elastic and in elastic collisionthe two bodies move in different directions after thecollision. the torque acting on mass m1 is = I * or w * l = I * or = (w * l/I) w = m1 * g,l = 0.30 m and I =(m1 * l2/3) or = (m1 * g * l/(m1 *l2/3)) = (3g/l) g = 9.8 m/s2 and l = 0.30 m the linear speed of mass m2 is v = r * w r = 0.050 m the collision between mass m1 and m2 iselastic therefore the momentum is conserved during elasticcollision (m1 + m3) * v = m2 * v or m3 = m2 - m1 = 1.2 - 1.0 =0.2 kg (ii)The recoil velocity of m1 is same as the recoilvelocity of m2.This is because the collision betweenm1 and m2 is elastic and in elastic collisionthe two bodies move in different directions after thecollision.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---