Question

In the system of two discs and a rod of mass 600 g each, a torque of magnitude \(43 \times 10^5\) dyne-cm is applied along the axis of rotation as shown in figure. Find the approx angular acceleration about given axis : 

• A. 11 rad/\(s^2\)
• B. 100 rad/\(s^2\)
• C. 27 rad/\(s^2\)
• D. 22 rad/\(s^2\)

Hint:

When using Parallel Axis Theorem for discs on a rod, remember to check if the axis is parallel to the diameter or the central axis of the disc. Here, the drawing indicates it's parallel to the diameter.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Angular acceleration \(\alpha\) is found using Newton's second law for rotation: \(\tau = I \alpha\).
The total moment of inertia \(I\) is the sum of the moments of inertia of the two discs and the rod about the specific axis.
Step 2: Key Formula or Approach:
1. \(\tau = I \alpha\).
2. Moment of inertia of a disc about its diameter: \(\frac{1}{4}MR^2\).
3. Moment of inertia of a rod about an axis at distance \(x\) from center: \(I_{rod} = \frac{1}{12}ML^2 + Mx^2\).
Step 3: Detailed Explanation:
Data (CGS): \(M = 600 \text{ g}\), \(R = 10 \text{ cm}\), Rod length \(L = 30 \text{ cm}\).
The axis is \(10 \text{ cm}\) from one end and \(20 \text{ cm}\) from the other.
Center of rod is at \(15 \text{ cm}\) from ends. Distance from center to axis \(x = 5 \text{ cm}\).
1. Rod: \(I_{rod} = \frac{1}{12}(600)(30^2) + 600(5^2) = 50(900) + 600(25) = 45000 + 15000 = 60000 \text{ g}\cdot\text{cm}^2\).
2. Disc 1 (at \(10 \text{ cm}\)): \(I_{d1} = \frac{1}{4}(600)(10^2) + 600(10^2) = 15000 + 60000 = 75000 \text{ g}\cdot\text{cm}^2\).
3. Disc 2 (at \(20 \text{ cm}\)): \(I_{d2} = \frac{1}{4}(600)(10^2) + 600(20^2) = 15000 + 240000 = 255000 \text{ g}\cdot\text{cm}^2\).
Total \(I = 60000 + 75000 + 255000 = 390000 \text{ g}\cdot\text{cm}^2 = 3.9 \times 10^5 \text{ g}\cdot\text{cm}^2\).
Given torque \(\tau = 43 \times 10^5 \text{ dyne-cm}\).
\[ \alpha = \frac{\tau}{I} = \frac{43 \times 10^5}{3.9 \times 10^5} \approx 11.02 \text{ rad/s}^2 \]
Step 4: Final Answer:
The approximate angular acceleration is 11 rad/\(s^2\).