Question

Problem 9 m,r A solid ball of mass m and radius r sits at rest at the top of a hill of height H leading to a circular loop-the loop. The center of mass of the ball will move in a circle of radius R if it goes around the loop. The moment of inertia of a solid ball is Ibull--mr. (a) Find an expression for the minimum height H for which the ball barely goes around the loop, staying in contact with the track all the way around, assuming that it rolls without slipping. (b) How does your answer relate to the minimum height for the earlier homework problem where it was a block that slid around a frictionless track? (That minimum height was 2.5 R.) Does this new answer make sense? If it is higher, where did the extra potential energy go? If it is lower, where did the extra kinetic energy come from? (c) The assumption that the ball will roll around the track without slipping if released from this estimated minimum height is not a good one. Why not? Hint: in order to keep rolling, there must bea force of static friction. Is there anywhere along its path where the force of static friction drops to zero?

Answer 1

m.r

a)

v_b = minimum speed at B for looping the loop = sqrt(5gR)

w_b = angular speed at B = v_b /r = sqrt(5gR)/r

H = height from where the ball is released

I_ball = moment of inertia of the ball = 2 mr²/5

Using conservation of energy between point A and B

Total Energy at A = Total energy at B

mg H = (0.5) I_ball w_b² + (0.5) m v_b²

mg H = (0.5) (2 mr²/5) (sqrt(5gR)/r)² + (0.5) m (sqrt(5gR))²

mg H = (mr²/5) (5gR/r²) + (0.5) m (5gR)

mg H = (m) (gR) + (0.5) m (5gR)

H = R + (0.5) (5) R

H = 3.5 R

b)

The minimum height is larger now as compared to the case when block was sliding.

yes the answer make sense.because there is rotational kinetic energy and kinetic energy involved. so greater amount of potential energy was required.

The extra potential energy goes into rotational kinetic energy of the ball.