Question

The masses of a solid cylinder and a hollow cylinder are 3.2 kg and 1.6 kg respectively. Both the solid cylinder and hollow cylinder start from rest from the top of an inclined plane and roll down without slipping. If both the cylinders have equal radius and the acceleration of the solid cylinder is \( 4 { ms}^{-2} \), the acceleration of the hollow cylinder is:

• A. \( 2 { ms}^{-2} \)
• B. \( 9 { ms}^{-2} \)
• C. \( 6 { ms}^{-2} \)
• D. \( 3 { ms}^{-2} \)

Hint:

For rolling motion, the acceleration down an incline depends on the moment of inertia. Use \( a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}} \) to compare different rolling objects.

Answer 1

The Correct Option is D

Step 1: Acceleration of rolling objects
The acceleration of a rolling object down an incline is given by: \[ a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}. \] where: - \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of inclination,
- \( K \) is the radius of gyration,
- \( R \) is the radius of the cylinder.
Step 2: Moment of inertia considerations
For different objects:
- Solid Cylinder: \( I = \frac{1}{2} m R^2 \), so \( K^2 = \frac{R^2}{2} \).
- Hollow Cylinder: \( I = m R^2 \), so \( K^2 = R^2 \).
Step 3: Acceleration ratio
The acceleration expression becomes: - For the solid cylinder: \[ a_s = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}}. \] \[ a_s = \frac{2}{3} g \sin \theta. \] - For the hollow cylinder: \[ a_h = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}. \] Step 4: Finding \( a_h \)
Given \( a_s = 4 { ms}^{-2} \), we set up the ratio: \[ \frac{a_h}{a_s} = \frac{\frac{g \sin \theta}{2}}{\frac{2 g \sin \theta}{3}} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}. \] \[ a_h = \frac{3}{4} \times 4 = 3 { ms}^{-2}. \] Step 5: Conclusion
Thus, the acceleration of the hollow cylinder is: \[ 3 { ms}^{-2}. \]

Answer 2

The problem is determining the acceleration of a hollow cylinder rolling down an inclined plane without slipping. We know the acceleration of a solid cylinder, which has a mass of 3.2 kg, is \(4 \, \text{ms}^{-2}\). The mass of the hollow cylinder is 1.6 kg. Both cylinders have the same radius.

For a cylinder rolling without slipping, the acceleration \(a\) is determined by the formula:

\(a = \frac{g \cdot \sin(\theta)}{1 + \frac{I}{m r^2}}\)

where \(I\) is the moment of inertia, \(m\) is the mass, and \(r\) is the radius of the cylinder.

• For a solid cylinder, \(I = \frac{1}{2}m r^2\).
• For a hollow cylinder, \(I = m r^2\).

Step 1: Calculate acceleration for a solid cylinder

The known acceleration for the solid cylinder is \(4 \, \text{ms}^{-2}\). Plug into the formula:

\(4 = \frac{g \cdot \sin(\theta)}{1 + \frac{1}{2}}\)

Simplify:

\(4 = \frac{g \cdot \sin(\theta)}{1.5} \Rightarrow g \cdot \sin(\theta) = 6\)

Step 2: Calculate acceleration for the hollow cylinder

Plug this into the formula for the hollow cylinder:

\(a_\text{hollow} = \frac{g \cdot \sin(\theta)}{1 + 1}\)

Given \(g \cdot \sin(\theta) = 6\):

\(a_\text{hollow} = \frac{6}{2} = 3 \, \text{ms}^{-2}\)

Thus, the acceleration of the hollow cylinder is 3 ms\(^{-2}\).