Answer:
To solve these problems, we can use the equations of rotational motion. The key equation we will use relates angular acceleration, initial angular velocity, final angular velocity, and time:
\[ \omega_f = \omega_i + \alpha t \]
where:
- \( \omega_f \) is the final angular velocity,
- \( \omega_i \) is the initial angular velocity,
- \( \alpha \) is the angular acceleration, and
- \( t \) is the time.
Given:
- \( \alpha = 4.44 \) rad/s²,
- \( \omega_i = 0 \) (initially at rest),
- \( \omega_f = 1.50 \) rad/s (for part a),
- \( r = 1.35 \) m (for part c).
### Part (a)
We want to find the time taken to reach a final angular velocity \( \omega_f = 1.50 \) rad/s.
\[ \omega_f = \omega_i + \alpha t \]
\[ 1.50 = 0 + 4.44 \cdot t \]
\[ t = \frac{1.50}{4.44} \]
### Part (b)
To find the number of revolutions, we need to convert the final angular velocity to revolutions per second:
\[ \text{Revolutions per second} = \frac{\omega_f}{2\pi} \]
Then, the total number of revolutions is the revolutions per second multiplied by the time \( t \) found in part (a).
### Part (c)
To find the time taken to stop, we can use the equation:
\[ F \cdot r = I \cdot \alpha \]
where \( F \) is the force, \( r \) is the radius, and \( I \) is the moment of inertia. Given \( F = 300 \) N, \( r = 1.35 \) m, and \( \alpha = -4.44 \) rad/s² (negative because it's a slowing force):
\[ 300 \cdot 1.35 = I \cdot (-4.44) \]
We can calculate \( I \) using \( I = \frac{1}{2} M R^2 \), where \( M \) is the mass of the merry-go-round. If you provide the mass or the moment of inertia, I can calculate the time to stop.
Explanation:
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