Equal torques asct on the discs A and B of theh previous problem, initially both being at rest. At a later instant, theliear speeds of a point on therim of a A another potin on the rim of B are `V_A and V_B` respectively. We have

Options

A. `V_AltV_B`
B. `V_A=V_B`
C. `V_AltV_B`
D. `the relation depends on the tactual magnitude of the torques

Answer

To solve the problem, we need to establish the relationship between the linear speeds \( V_A \) and \( V_B \) of points on the rims of discs A and B, respectively. Here's a step-by-step solution:

### Step 1: Understand the Given Information
We have two discs, A and B:
- Disc A has a radius \( r \) and moment of inertia \( I_A \).
- Disc B has a radius \( 4r \) and moment of inertia \( I_B \).
- Equal torques \( \tau \) are applied to both discs, and they start from rest.

### Step 2: Relate the Moments of Inertia
From the previous problem, we know that:
\[ I_B = 64 I_A \]
This is because the moment of inertia for a disc is proportional to the square of its radius, and since \( 4r \) is four times the radius of \( r \), the moment of inertia increases by a factor of \( 4^2 = 16 \). However, since the thickness is also considered, the total factor becomes \( 16 \times 4 = 64 \).

### Step 3: Calculate Angular Acceleration
Using the relation for torque:
\[ \tau = I_A \alpha_A \quad \text{and} \quad \tau = I_B \alpha_B \]
We can express the angular accelerations \( \alpha_A \) and \( \alpha_B \):
\[ \alpha_A = \frac{\tau}{I_A} \quad \text{and} \quad \alpha_B = \frac{\tau}{I_B} \]
Substituting \( I_B = 64 I_A \):
\[ \alpha_B = \frac{\tau}{64 I_A} = \frac{1}{64} \frac{\tau}{I_A} = \frac{1}{64} \alpha_A \]

### Step 4: Calculate Angular Velocities
Since both discs start from rest, we can use the equation for angular velocity:
\[ \omega_A = \alpha_A t \quad \text{and} \quad \omega_B = \alpha_B t \]
Substituting for \( \alpha_B \):
\[ \omega_B = \left(\frac{1}{64} \alpha_A\right) t = \frac{1}{64} \omega_A \]

### Step 5: Relate Linear Velocities
The linear velocities at the rim of the discs are given by:
\[ V_A = r \omega_A \quad \text{and} \quad V_B = 4r \omega_B \]
Substituting \( \omega_B \):
\[ V_B = 4r \left(\frac{1}{64} \omega_A\right) = \frac{4r}{64} \omega_A = \frac{r}{16} \omega_A \]

### Step 6: Substitute for \( V_A \)
Now substituting for \( V_A \):
\[ V_A = r \omega_A \]
Thus, we can express the relationship between \( V_A \) and \( V_B \):
\[ V_B = \frac{r}{16} \frac{V_A}{r} = \frac{1}{16} V_A \]

### Step 7: Conclusion
From the above relationship, we conclude:
\[ V_A = 16 V_B \]
This means that the linear speed of point A is greater than that of point B, confirming that:
\[ V_A > V_B \]