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Step-by-Step Solution
Step 1: Write down the known quantities
• Radius of pulley, $R = 2\,\text{m}$.
• Moment of inertia of pulley, $I = 10\,\text{kg·m}^2$.
• Tangential force applied, $F(t) = 20t - 5t^2$ (in newtons).
• We want to find the number of rotations before the pulley reverses its direction.
Step 2: Express the torque on the pulley
The torque $\tau(t)$ due to the tangential force is given by
$$
\tau(t) = F(t)\,R = \bigl(20t - 5t^2\bigr)\times 2.
$$
Therefore,
$$
\tau(t) = 40t - 10t^2.
$$
Step 3: Find the angular acceleration
The angular acceleration $\alpha(t)$ is given by
$$
\alpha(t) = \frac{\tau(t)}{I}.
$$
Substituting $\tau(t)$ and $I$:
$$
\alpha(t) = \frac{40t - 10t^2}{10} = 4t - t^2.
$$
Thus,
$$
\alpha(t) = 4t - t^2.
$$
Step 4: Determine the angular velocity as a function of time
By definition,
$$
\alpha(t) = \frac{d\omega}{dt}.
$$
Hence,
$$
\frac{d\omega}{dt} = 4t - t^2.
$$
Integrate both sides from $0$ to $t$ (assuming the pulley starts from rest, so $\omega(0)=0$):
$$
\omega(t) = \int_{0}^{t} (4t' - t'^2)\,dt'.
$$
Let us use a dummy variable $t'$ for integration. Then
$$
\omega(t) = \left[\ 2t'^2 - \frac{t'^3}{3}\right]_{0}^{t}
= 2t^2 - \frac{t^3}{3}.
$$
Step 5: Find the time when the direction reverses
The direction reverses when $\omega(t)$ becomes zero again (after $t=0$). Solve
$$
2t^2 - \frac{t^3}{3} = 0,\quad t > 0.
$$
Factor out $t^2$:
$$
t^2 \Bigl(2 - \frac{t}{3}\Bigr) = 0.
$$
One solution is $t=0$ and the other comes from $2 - \frac{t}{3}=0$, giving $t=6\,\text{s}$.
Therefore, the pulley’s angular velocity returns to zero at $t = 6\,\text{s}$, reversing direction thereafter.
Step 6: Calculate the total angular displacement from $t=0$ to $t=6\,\text{s}$
The angular displacement $\theta$ is obtained by integrating $\omega(t)$:
$$
\theta = \int_{0}^{6} \omega(t)\,dt.
$$
Substituting $\omega(t)= 2t^2 - \frac{t^3}{3}$:
$$
\theta = \int_{0}^{6} \Bigl(2t^2 - \frac{t^3}{3}\Bigr)\,dt.
$$
Let us compute this integral step by step:
$$
\int 2t^2\,dt = \frac{2}{3}\,t^3,
$$
$$
\int \frac{t^3}{3}\,dt = \frac{t^4}{12}.
$$
Hence,
$$
\theta = \biggl[\frac{2}{3} t^3 - \frac{t^4}{12}\biggr]_{0}^{6}.
$$
Evaluate at $t=6$:
$$
\theta
= \left(\frac{2}{3} \times 6^3\right)
- \left(\frac{6^4}{12}\right)
= \frac{2}{3} \times 216 - \frac{1296}{12}.
$$
$$
= 144 - 108 = 36\,\text{rad}.
$$
Step 7: Convert the angular displacement to number of rotations
One complete rotation corresponds to $2\pi$ radians. Therefore, the number of rotations $n$ is
$$
n = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{36}{6.2832...} \approx 5.73.
$$
This value is more than 3 but less than 6.
Conclusion
The pulley makes around 5.73 rotations before its direction of motion is reversed. Hence, the answer
“more than 3 but less than 6” is correct.
