Question:
A metre scale is standing vertically on the earthâs surface on one of its end. It now falls on earth without slipping. Find the velocity with which the free end of the scale strikes the earth. (\( g = 10 \, \text{m/s}^2 \))
A metre scale is standing vertically on the earthâs surface on one of its end. It now falls on earth without slipping. Find the velocity with which the free end of the scale strikes the earth. (\( g = 10 \, \text{m/s}^2 \))
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For a rolling object, the total kinetic energy is the sum of translational and rotational kinetic energy.
Updated On: Apr 22, 2026
- 9.8 m/s
- 5.47 m/s
- 4.5 m/s
- 1 m/s
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The Correct Option is B
Solution and Explanation
Step 1: Understand the motion of the scale.
The scale falls without slipping, meaning it rotates about the point of contact with the ground. The motion is a combination of both rotational and translational motion. The velocity of the free end of the scale can be calculated using the energy conservation principle.
Step 2: Energy conservation approach.
Initially, the potential energy of the scale is given by \( U = mgh \), where \( h = 1 \, \text{m} \) (height of the scale). As the scale falls, this potential energy is converted into both translational and rotational kinetic energy. The total energy at the moment just before it strikes the ground is: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 \] where \( v \) is the velocity of the center of mass, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. Using the relationship between the linear velocity and angular velocity for rolling motion \( v = \omega r \), the equation simplifies.
Step 3: Final velocity calculation.
Using the rotational dynamics equations and energy conservation, we can find that the velocity of the free end when it strikes the ground is approximately: \[ v = 5.47 \, \text{m/s} \]
Step 4: Conclusion.
The velocity with which the free end of the scale strikes the earth is \( 5.47 \, \text{m/s} \), which is option (2).
The scale falls without slipping, meaning it rotates about the point of contact with the ground. The motion is a combination of both rotational and translational motion. The velocity of the free end of the scale can be calculated using the energy conservation principle.
Step 2: Energy conservation approach.
Initially, the potential energy of the scale is given by \( U = mgh \), where \( h = 1 \, \text{m} \) (height of the scale). As the scale falls, this potential energy is converted into both translational and rotational kinetic energy. The total energy at the moment just before it strikes the ground is: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 \] where \( v \) is the velocity of the center of mass, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. Using the relationship between the linear velocity and angular velocity for rolling motion \( v = \omega r \), the equation simplifies.
Step 3: Final velocity calculation.
Using the rotational dynamics equations and energy conservation, we can find that the velocity of the free end when it strikes the ground is approximately: \[ v = 5.47 \, \text{m/s} \]
Step 4: Conclusion.
The velocity with which the free end of the scale strikes the earth is \( 5.47 \, \text{m/s} \), which is option (2).
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