Question

Two flywheels are connected by a non-slipping belt as shown in the figure. \(m_1 = 4 \, \text{kg}\), \(r_1 = 20 \, \text{cm}\), \(m_2 = 20 \, \text{kg}\), \(r_2 = 30 \, \text{cm}\). A torque of 10 Nm is applied on the smaller wheel. Then match the entries of column I with appropriate entries of column II. 

• A. \( a \rightarrow iii, b \rightarrow i, c \rightarrow ii \)
• B. \( a \rightarrow iii, b \rightarrow ii, c \rightarrow i \)
• C. \( a \rightarrow ii, b \rightarrow iii, c \rightarrow i \)
• D. \( a \rightarrow ii, b \rightarrow iii, c \rightarrow ii \)

Hint:

When dealing with non-slipping belts, remember the relation between angular accelerations: \( \alpha_1 r_1 = \alpha_2 r_2 \). Also, to find torque and angular acceleration, use the moment of inertia and apply the appropriate equations.

Answer 1

The Correct Option is B

Given the conditions of the problem, we know that the flywheels are connected by a non-slipping belt. Therefore, the angular acceleration of the smaller wheel (\( \alpha_1 \)) and the larger wheel (\( \alpha_2 \)) are related by the following equation: \[ \alpha_1 r_1 = \alpha_2 r_2 \] where \( r_1 \) and \( r_2 \) are the radii of the smaller and larger wheels, respectively. 1. Angular acceleration of the smaller wheel (\( \alpha_1 \)): The angular acceleration of the smaller wheel is determined from the applied torque \( \tau_1 \) using the relation: \[ \tau_1 = I_1 \alpha_1 \] where \( I_1 \) is the moment of inertia of the smaller wheel, given by \( I_1 = \frac{1}{2} m_1 r_1^2 \). Thus: \[ I_1 = \frac{1}{2} \times 4 \, \text{kg} \times (0.2 \, \text{m})^2 = 0.08 \, \text{kg} \, \text{m}^2 \] The torque applied is \( \tau_1 = 10 \, \text{Nm} \), so: \[ \alpha_1 = \frac{\tau_1}{I_1} = \frac{10}{0.08} = 125 \, \text{rad/s}^2 \] 2. Angular acceleration of the larger wheel (\( \alpha_2 \)): Using the relationship \( \alpha_1 r_1 = \alpha_2 r_2 \), we can find \( \alpha_2 \): \[ \alpha_2 = \frac{\alpha_1 r_1}{r_2} = \frac{125 \times 0.2}{0.3} = 83.33 \, \text{rad/s}^2 \] 3. Torque on the larger wheel (\( \tau_2 \)): The torque on the larger wheel is given by: \[ \tau_2 = I_2 \alpha_2 \] where \( I_2 = \frac{1}{2} m_2 r_2^2 \). Thus: \[ I_2 = \frac{1}{2} \times 20 \, \text{kg} \times (0.3 \, \text{m})^2 = 0.9 \, \text{kg} \, \text{m}^2 \] Therefore: \[ \tau_2 = 0.9 \times 83.33 = 75 \, \text{Nm} \] Thus, the numerical values corresponding to the quantities are: - \( \alpha_1 = 125 \, \text{rad/s}^2 \) (which matches option iii), - \( \tau_2 = 75 \, \text{Nm} \) (which matches option ii), - \( \alpha_2 = 83.33 \, \text{rad/s}^2 \) (which matches option i).