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Tomtit [17]
3 years ago
7

A machine part has the shape of a solid uniform sphere of mass 245 g and diameter 4.30 cm . It is spinning about a frictionless

axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.
Physics
1 answer:
Alla [95]3 years ago
3 0

Answer:

Angular acceleration, \alpha =9.49\ rad/s^2

Explanation:

It is given that,

Mass of the solid sphere, m = 245 g = 0.245 kg

Diameter of the sphere, d = 4.3 cm = 0.043 m

Radius, r = 0.0215 m

Force acting at a point, F = 0.02 N

Let \alpha is its angular acceleration. The relation between the angular acceleration and the torque is given by :

\tau=I\times \alpha

I is the moment of inertia of the solid sphere

For a solid sphere, I=\dfrac{2}{5}mr^2

\alpha =\dfrac{\tau}{I}

\alpha =\dfrac{F.r}{(2/5)mr^2}

\alpha =\dfrac{5F}{2mr}

\alpha =\dfrac{5\times 0.02}{2\times 0.245\times 0.0215}

\alpha =9.49\ rad/s^2

So, its angular acceleration is 9.49\ rad/s^2. Hence, this is the required solution.

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