A wheel ios rolling without sliding on a horizontal surface. The centre of the wheelk moves with a constant speed `v_(0)`. Consider a point P on the rim which is at the top at time t=0. The square of speed of point P is plooted against time t. The correct plot is (R is radius of the wheel)

Options

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Answer

To solve the problem of a wheel rolling without sliding on a horizontal surface, we need to analyze the motion of a point on the rim of the wheel. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the motion of the wheel
The wheel rolls without slipping, which means that the point of contact with the ground is momentarily at rest. The center of the wheel moves with a constant speed \( v_0 \).

### Step 2: Identify the position of point P
At time \( t = 0 \), point P is at the top of the wheel. As the wheel rolls, point P will move in a circular path around the center of the wheel.

### Step 3: Determine the speed of point P
The speed of point P can be expressed as the sum of the translational speed of the center of the wheel and the rotational speed due to the wheel's rotation.

- The translational speed of the center is \( v_0 \).
- The rotational speed at the top of the wheel is \( v_{\text{rot}} = R \omega \), where \( R \) is the radius of the wheel and \( \omega \) is the angular velocity.

Since the wheel rolls without slipping, we have the relationship:
\[ v_0 = R \omega \]
Thus, the speed of point P at any time \( t \) can be expressed as:
\[ v_P(t) = v_0 + R \omega(t) \]

### Step 4: Express angular velocity in terms of time
The angular position \( \theta \) of the wheel as a function of time can be expressed as:
\[ \theta(t) = \omega t \]
The angular velocity \( \omega \) is constant, so:
\[ \omega = \frac{v_0}{R} \]

### Step 5: Calculate the speed of point P
At time \( t \), the speed of point P becomes:
\[ v_P(t) = v_0 + R \left( \frac{v_0}{R} \sin(\omega t) \right) \]
This simplifies to:
\[ v_P(t) = v_0 (1 + \sin(\omega t)) \]

### Step 6: Find the square of the speed
To plot the square of the speed against time:
\[ v_P^2(t) = \left( v_0 (1 + \sin(\omega t)) \right)^2 \]
Expanding this gives:
\[ v_P^2(t) = v_0^2 (1 + 2\sin(\omega t) + \sin^2(\omega t)) \]

### Step 7: Analyze the behavior of \( v_P^2(t) \)
The term \( \sin^2(\omega t) \) oscillates between 0 and 1, and thus the square of the speed will oscillate between:
\[ v_0^2 (1 + 0) \text{ and } v_0^2 (1 + 2 + 1) = 4v_0^2 \]
This indicates that \( v_P^2(t) \) will have a periodic behavior with a maximum and minimum value.

### Step 8: Identify the correct plot
Given that the square of the speed oscillates sinusoidally, the correct plot will show a periodic function that varies between \( v_0^2 \) and \( 4v_0^2 \).

### Conclusion
The correct plot of the square of the speed of point P against time \( t \) is a sinusoidal curve oscillating between \( v_0^2 \) and \( 4v_0^2 \).

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