Question

A solid cylinder of mass 2 kg, length 40 cm and radius 10 cm is placed in contact with a solid sphere of mass 0.5 kg and radius 10 cm such that the centres of the two bodies lie along the geometrical axis of the cylinder. The distance of the centre of mass of the system of two bodies from the centre of the sphere is

• A. 27 cm
• B. 15 cm
• C. 24 cm
• D. 18 cm

Hint:

Drawing a simple 1D diagram of the setup is extremely helpful. Placing the origin at the location of one of the objects often simplifies the calculation, as one of the position terms (\(x_1\)) becomes zero.

Answer 1

The Correct Option is C

Step 1: Define the Coordinate System: Let the center of the sphere be the origin \( (0, 0) \). The centers of both bodies lie on the X-axis (geometrical axis).
Step 2: Identify Positions of Centers of Mass: Sphere: Mass \( m_1 = 0.5 \, \text{kg} \). Radius \( R_{sphere} = 10 \, \text{cm} \). Center of mass position \( x_1 = 0 \) (since it's at the origin). Cylinder: Mass \( m_2 = 2 \, \text{kg} \). Radius \( R_{cyl} = 10 \, \text{cm} \). Length \( L = 40 \, \text{cm} \). The cylinder is placed "in contact" with the sphere along the axis. The flat face of the cylinder touches the sphere? Or the curved surface? The problem says "centres... lie along the geometrical axis". This usually implies they are placed end-to-end. Distance from origin (sphere center) to the contact point = Radius of sphere = 10 cm. Distance from the contact point to the center of the cylinder = Half the length of the cylinder = \( L/2 = 40/2 = 20 \, \text{cm} \). Position of cylinder's center of mass \( x_2 = 10 \, \text{cm} + 20 \, \text{cm} = 30 \, \text{cm} \).
Step 3: Calculate System Center of Mass (\( X_{CM} \)): \[ X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \] \[ X_{CM} = \frac{0.5(0) + 2(30)}{0.5 + 2} \] \[ X_{CM} = \frac{60}{2.5} \] \[ X_{CM} = \frac{600}{25} = 24 \, \text{cm} \]