A solid, uniform spherical boulder starts from rest and rolls down a 50.0-m high

Question

A solid, uniform spherical boulder starts from rest and rolls down a 50.0-m high hill, as shown in the figure (Figure 1) . The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction.

What is the translational speed of the boulder when it reaches the bottom of the hill?

A solid, uniform spherical boulder starts from rest and rolls down a 50.0-m high hill, as shown in the figure (Figure 1) . The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill?

Explanation / Answer

h/2 = 25 m.

KE added by each half = mgh/2. In

the first (rolling) half, KE is divided between translational and rotational KE, with 2/7 going to rotational and 5/7 to translational.

In the second (sliding) half all the added KE is translational.
Total translational KE,

TKE = (mgh/2)*(5/7 + 1) = 6mgh/7

v = sqrt(2TKE/m)
Combining to eliminate m,

v = sqrt(12gh/7)

v= 28.99 m/s

v=29m/s

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