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A spherical boulder of mass 98.1 kg and radius 22 cm rolls without slipping down a hill 13 m high from rest.

(a) What is its angular momentum about its center when it is halfway down the hill?

Ans: 82.4 kg·m²/s

Answer :

Final answer:

The angular momentum of a rolling, spherical boulder can be calculated using the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. By calculating the moment of inertia and angular velocity, we can determine the angular momentum when the boulder is half way down the hill.

Explanation:

The angular momentum of a spherical boulder rolling without slipping can be calculated using the formula:

L = Iω

Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

For a solid sphere, the moment of inertia is given by:

I = (2/5)mR^2

Where m is the mass of the sphere and R is the radius.

Given the mass and radius of the boulder, we can calculate its moment of inertia.

Since the boulder is rolling without slipping, the linear velocity at the center of the boulder is equal to ωR. We can use this relationship to calculate the angular velocity.

Finally, we can calculate the angular momentum when the boulder is half way down the hill.

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