Question

A body of mass 0.2 kg is attached to a light string of length 1 m and revolved in a vertical circle. What is the minimum speed at the lowest point so that the body can complete the circular motion? (Take \( g = 10 \, \text{m/s}^2 \))

• A. 2 m/s
• B. 3.16 m/s
• C. 5 m/s
• D. 6.32 m/s

Hint:

Remember: When applying energy principles, the minimum speed corresponds to the condition when the tension in the string is zero at the lowest point.

Answer 1

The Correct Option is B

Minimum Velocity for Circular Trajectory

Provided data:

• Object mass \( m = 0.2 \, \text{kg} \)
• String length \( L = 1 \, \text{m} \)
• Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)

Step 1: Define conditions for circular motion.

For continuous circular motion, the string tension at the apex must be non-negative. At the apex, forces on the object are gravity \( mg \) and tension \( T \), both contributing to the centripetal force.

At the nadir, sufficient velocity is required to sustain circular motion. The minimum velocity occurs when tension at the apex is zero, meaning gravitational force alone provides the centripetal force there.

Step 2: Apply centripetal force equation at the apex.

The centripetal force at the apex is given by:

\[ \frac{mv^2}{L} = mg \] where \( v \) is velocity at the apex, \( m \) is mass, \( L \) is string length (radius), and \( g \) is gravitational acceleration.

Step 3: Calculate minimum velocity.

Solving for \( v \) (minimum velocity at the nadir):

\[ \frac{mv^2}{L} = mg \quad \Rightarrow \quad v^2 = gL \quad \Rightarrow \quad v = \sqrt{gL} \]

Substituting values:

\[ v = \sqrt{10 \times 1} = \sqrt{10} \approx 3.16 \, \text{m/s} \]

Outcome:

The minimum velocity at the nadir required for the object to complete circular motion is approximately \( 3.16 \, \text{m/s} \).