A man rolls a large spherical rock ( m = 34,000 kg, R = 1.6 m) down a hill towar

Question

A man rolls a large spherical rock (m = 34,000 kg, R = 1.6 m) down a hill towards his enemy. The hill is inclined at an angle = 34o above horizontal, and the rock starts from rest at a height H= 55 m above the enemy. The rock rolls without sliding. What is the acceleration of the rock as it rolls down the hill? You must answer symbolically in terms of m, g, R, H, and/or theta

Correct Answer: (5/7)*g*sin(theta)

16. [4pt]
You do not need to answer the previous question to solve this one. In the previous problem, when the rock reaches the enemy, what fraction of the total kinetic energy is translational rather than rotational? Your answer is a positive number between 0 and 1.

ONLY NEED #16 (correct answer is .714, but I don't know how to do it)

Explanation / Answer

here, it is case of pure rolling. so angular acceleration alpha = a/r

balancing torque about contact point,

torque = mg [sin theta ]*r

i* alpha = mg [sin theta ]*r^2 where i = 2/5 mr^2 + mr^2 = 7/5mr^2

  7/5mr^2 * alpha = = mg [sin theta ]*r^2

alpha = 5/7 *g sin theta answer

16] translational KE= 0.5mv^2

rotational KE= 0.5iw^2 = 0.5 *(2/5 mr^2) w^2 because i=2/5mr^2 and w=v/r

= 0.2 mv^2

translational KE/Total KE = 0.5mv^2/(0.5mv^2+ 0.2 mv^2)

= 5/7 = 0.714 answer

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