To find the kinetic energy of the merry-go-round after 3.25 seconds, we will follow these steps:
- Calculate the moment of inertia (\( I \)) of the solid cylinder (merry-go-round).
- Determine the angular acceleration (\( \alpha \)) produced by the tangential force.
- Calculate the angular velocity (\( \omega \)) after 3.25 seconds.
- Finally, compute the kinetic energy (\( KE \)) of the merry-go-round.
Step 1: Moment of Inertia
The moment of inertia \( I \) of a solid cylinder about its central axis is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
Where:
- \( m \) is the mass of the cylinder
- \( r \) is the radius of the cylinder
First, we need to find \( m \) (the mass of the merry-go-round). We can find it using the weight:
\[ W = m g \quad \Rightarrow \quad m = \frac{W}{g} = \frac{661 , \text{N}}{9.8 , \text{m/s}^2} \approx 67.4 , \text{kg} \]
Now substituting the values into the moment of inertia formula:
\[ I = \frac{1}{2} (67.4 , \text{kg}) (1.01 , \text{m})^2 \approx \frac{1}{2} (67.4) (1.0201) \approx 34.4 , \text{kg m}^2 \]
Step 2: Angular Acceleration
The torque (\( \tau \)) due to the tangential force is given by:
\[ \tau = F \cdot r = 75.4 , \text{N} \cdot 1.01 , \text{m} \approx 76.054 , \text{N m} \]
Using Newton's second law for rotation (\( \tau = I \alpha \)), we can find the angular acceleration \( \alpha \):
\[ \alpha = \frac{\tau}{I} = \frac{76.054 , \text{N m}}{34.4 , \text{kg m}^2} \approx 2.21 , \text{rad/s}^2 \]
Step 3: Angular Velocity
Since the merry-go-round starts from rest, the angular velocity \( \omega \) after \( t = 3.25 \) seconds can be calculated using the formula:
\[ \omega = \alpha t = 2.21 , \text{rad/s}^2 \cdot 3.25 , \text{s} \approx 7.18 , \text{rad/s} \]
Step 4: Kinetic Energy
The kinetic energy \( KE \) of a rotating object is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
Substituting in our values:
\[ KE = \frac{1}{2} (34.4 , \text{kg m}^2) (7.18 , \text{rad/s})^2 \approx \frac{1}{2} (34.4) (51.6324) \approx 889.048 , \text{J} \]
Finally, rounding to three significant figures, we find the kinetic energy of the merry-go-round after 3.25 seconds is approximately:
\[ \boxed{889 , \text{J}} \]