A uniform sphere with a mass of 23.0 kg and a radius of 0.330 m rotates at a constant angular velocity around a stationary axis that coincides with the sphere's diameter. Part A: If the sphere's kinetic energy is 238 J, what is the tangential velocity of a point on the sphere's rim?
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To find the tangential velocity of a point on the rim of the rotating sphere, we first need to know the relationship between the kinetic energy of a rotating object and its angular velocity. The kinetic energy (KE) for a rotating object can be expressed as:
\[ KE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the object and \( \omega \) is the angular velocity.
The moment of inertia for a solid sphere rotating about an axis through its center is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the sphere and \( r \) is its radius.
Given that the mass \( m \) of the sphere is 23.0 kg and the radius \( r \) is 0.330 m, let's calculate \( I \):
\[ I = \frac{2}{5} (23.0 \, \text{kg}) (0.330 \, \text{m})^2 = \frac{2}{5} (23.0) (0.1089) = 2 \times 4.6045 = 9.209 \, \text{kg} \cdot \text{m}^2 \]
Next, we'll rearrange the kinetic energy formula to solve for the angular velocity \( \omega \):
\[ \omega = \sqrt{\frac{2 KE}{I}} \]
Plugging in the given kinetic energy \( KE = 238 \) J and the calculated moment of inertia \( I \), we get:
\[ \omega = \sqrt{\frac{2 \times 238 \, \text{J}}{9.209 \, \text{kg} \cdot \text{m}^2}} = \sqrt{\frac{476}{9.209}} = \sqrt{51.694} \approx 7.191 \, \text{rad/s} \]
The tangential velocity \( v_t \) at the rim of the sphere is related to the angular velocity by:
\[ v_t = \omega r \]
Substituting the values for \( \omega \) and \( r \), we find:
\[ v_t = (7.191 \, \text{rad/s}) (0.330 \, \text{m}) \approx 2.373 \, \text{m/s} \]
So, the tangential velocity of a point on the sphere's rim is approximately 2.373 m/s.