A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t^2) N (where, t is measured

Q: Why do we set $\omega = 0$ to find the reversal point?

When an object reverses its direction of rotation, it must momentarily come to rest before spinning the other way. This means its angular velocity $\omega$ must pass through zero.

Q: Why did we ignore $t = 0$ when solving for the reversal time?

At $t = 0$, the pulley is just starting its motion from rest. It hasn't rotated yet, so it's not 'reversing' its direction at that instant. The reversal happens at the later time $t = 6\text{ s}$.

Q: Can we use the standard equations of motion like $\omega = \omega_0 + \alpha t$?

No! Those equations are strictly for *constant* angular acceleration. Here, our force (and thus our angular acceleration) depends on time ($t$), so we must use calculus (integration).

Q: Why do we divide the total angle by $2\pi$?

The angular displacement $\theta$ we calculated is in radians. Since one complete rotation corresponds to $2\pi$ radians, dividing the total radians by $2\pi$ gives the total number of full rotations.

Q: What does a negative torque mean in this context?

After $t = 4\text{ s}$, the force $F = 20t - 5t^2$ becomes negative, meaning it starts pushing in the opposite direction. This negative torque slows the pulley down until it stops and reverses at $t = 6\text{ s}$.

Question

A pulley of radius $2	ext{ m}$ is rotated about its axis by a force $F = (20t - 5t^2)	ext{ N}$ (where, $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10	ext{ kg-m}^2$, then the number of rotations made by the pulley before its direction of motion is reserved, is

Options:

(A) more than 3 but less than 6

(B) more than 6 but less than 9

(C) more than 9

(D) less than 3

Sigma Percentile · Accepted Answer

### The Setup: A Time-Varying Force

Imagine a heavy pulley, initially at rest, waiting to be spun. We are given its radius $R = 2	ext{ m}$ and its moment of inertia $I = 10	ext{ kg-m}^2$.

Suddenly, a tangential force starts acting on its edge. But this isn't your standard constant force; it's a dynamic, time-varying force described by the equation $F = (20t - 5t^2)	ext{ N}$.

This force will create a torque, causing the pulley to accelerate, spin, and eventually, as the force changes direction, slow down and reverse. Our mission is to find out exactly how many rotations it completes before that reversal happens.

### From Force to Angular Acceleration

To understand the pulley's rotational journey, we first need to find its angular acceleration, $\alpha$.

We rely on **Newton's Second Law for Rotation**, which states that the net torque $	au$ equals the moment of inertia $I$ times the angular acceleration $\alpha$.

Since the force is applied tangentially at the edge of the pulley, the torque is simply the force multiplied by the radius:
$$ 	au = F \cdot R $$

Equating this to $I\alpha$, we get:
$$ \alpha = \frac{F \cdot R}{I} $$

Let's substitute our given values into this master equation:
$$ \alpha = \frac{(20t - 5t^2) \cdot 2}{10} $$

Simplifying the fraction, the $2$ and $10$ reduce to a $5$ in the denominator. Dividing each term by $5$, we find our angular acceleration as a function of time:
$$ \alpha(t) = 4t - t^2 $$

### The Journey to Angular Velocity

Now that we have the acceleration, we need to find the angular velocity, $\omega$.

By definition, angular acceleration is the rate of change of angular velocity ($\alpha = \frac{d\omega}{dt}$). Therefore, to find $\omega(t)$, we must integrate $\alpha(t)$ with respect to time:
$$ \omega(t) = \int_0^t (4t - t^2) dt $$

Performing the integration, the $4t$ becomes $2t^2$, and the $t^2$ becomes $\frac{t^3}{3}$.

This gives us the equation for the pulley's angular velocity at any given moment:
$$ \omega(t) = 2t^2 - \frac{t^3}{3} $$

### Finding the Reversal Point

The problem asks for the number of rotations *before* the direction of motion is reversed.

What does a reversal look like physically? Just like a ball thrown straight up must momentarily stop before falling back down, the pulley must come to a complete stop for a split second before spinning the other way.

This means the reversal occurs exactly when the angular velocity is zero:
$$ \omega(t) = 0 $$

Setting our velocity equation to zero:
$$ 2t^2 - \frac{t^3}{3} = 0 $$

We can factor out $t^2$:
$$ t^2 \left(2 - \frac{t}{3}ight) = 0 $$

This yields two solutions: $t = 0$ (the moment it started) and $t = 6	ext{ s}$. The pulley reverses its direction exactly at the $6$-second mark!

### The Final Count: Angular Displacement and Rotations

We know the time of reversal, but we need the total angle turned, $	heta$.

Since angular velocity is the rate of change of angular displacement ($\omega = \frac{d	heta}{dt}$), we integrate $\omega(t)$ from $t = 0$ to $t = 6	ext{ s}$:
$$ 	heta = \int_0^6 \left(2t^2 - \frac{t^3}{3}ight) dt $$

Let's execute the integration:
$$ 	heta = \left[ \frac{2t^3}{3} - \frac{t^4}{12} ight]_0^6 $$

Plugging in the upper limit of $6$:
$$ 	heta = \frac{2(216)}{3} - \frac{1296}{12} $$
$$ 	heta = 144 - 108 = 36	ext{ rad} $$

The pulley has turned through a total of $36	ext{ radians}$. But the question asks for the number of *rotations*.

Since one full rotation is $2\pi	ext{ radians}$, we divide our total angle by $2\pi$:
$$ n = \frac{36}{2\pi} \approx \frac{36}{6.28} \approx 5.73	ext{ rotations} $$

Looking at our options, $5.73$ is clearly **more than 3 but less than 6**. The mystery of the reversing pulley is solved!

Select Answer:

Visualized Solution

$Physical Setup$

$Torque and Angular Acceleration$

$Substituting Values$

$Angular Acceleration α (t)$

$Angular Velocity ω (t)$

$Integrating for ω$

$Condition for Reversal$

$Time of Reversal$

$Angular Displacement θ$

$Calculating θ$

$Number of Rotations$

The Sigma Insight: Kinematics of Rotational Motion

The Setup

From Force to Angular Acceleration

The Journey to Angular Velocity

Finding the Reversal Point

The Final Count

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque and Angular Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Acceleration α(t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Velocity ω(t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating for ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Reversal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time of Reversal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Displacement θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Rotations