A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of the wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, will be:
Show Hint
- 20 rad/s
- 30 rad/s
- 10 rad/s
- 0 rad/s
The Correct Option is A
Approach Solution - 1
The work done \( W_f \) by the force \( F = 20 \, N \) is given by: \[ W_f = F \cdot d = 20 \times 1 = 20 \, J \] This is the change in kinetic energy of the wheel: \[ KE = \frac{1}{2} I \omega^2 \] Using \( I = MR^2 \) where \( M = 10 \, \text{kg} \) and \( R = 0.1 \, \text{m} \): \[ I = 10 \times (0.1)^2 = 0.1 \, \text{kg m}^2 \] Now equating the work done to the change in kinetic energy: \[ 20 = \frac{1}{2} \times 0.1 \times \omega^2 \quad \Rightarrow \quad \omega = 20 \, \text{rad/s} \]
Approach Solution -2
This problem involves calculating the final angular velocity of a wheel after a constant force is applied to a cord wrapped around its rim. The wheel is initially at rest and is treated as a ring since its spokes have negligible mass.
Concept Used:
The solution is based on the Work-Energy Theorem for rotational motion. This theorem states that the work done on a rigid body is equal to the change in its rotational kinetic energy.
The key formulas used are:
- Work Done (W): The work done by a constant force \( F \) applied over a distance \( L \) is \( W = F \cdot L \).
- Rotational Kinetic Energy (KE_rot): The kinetic energy of a rotating body is \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
- Moment of Inertia of a Ring (I): Since the mass of the wheel is concentrated on the rim, it is treated as a ring. The moment of inertia of a ring about its central axis is \( I = MR^2 \), where \( M \) is the mass and \( R \) is the radius.
- Work-Energy Theorem: \( W = \Delta KE_{rot} = KE_{rot, final} - KE_{rot, initial} \).
Step-by-Step Solution:
Step 1: List the given quantities in SI units.
The given parameters for the wheel and the applied force are:
- Mass of the wheel, \( M = 10 \, \text{kg} \)
- Radius of the wheel, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Applied force (pull), \( F = 20 \, \text{N} \)
- Length of the unwound cord, \( L = 1 \, \text{m} \)
- Initial angular velocity, \( \omega_i = 0 \, \text{rad/s} \) (since it starts from rest)
Step 2: Calculate the moment of inertia (\( I \)) of the wheel.
The wheel is considered a ring because its spokes have negligible mass, meaning all its mass is at the rim. The formula for the moment of inertia of a ring is:
\[ I = MR^2 \]Substituting the given values:
\[ I = (10 \, \text{kg}) \times (0.1 \, \text{m})^2 = 10 \times 0.01 = 0.1 \, \text{kg} \cdot \text{m}^2 \]Step 3: Calculate the work done (\( W \)) by the applied force.
The work is done by pulling the cord over a distance \( L \). The formula for work done is:
\[ W = F \times L \]Substituting the given values:
\[ W = (20 \, \text{N}) \times (1 \, \text{m}) = 20 \, \text{J} \]Step 4: Apply the Work-Energy Theorem.
The work done on the wheel is converted into its rotational kinetic energy. The theorem states:
\[ W = KE_{rot, final} - KE_{rot, initial} \]Since the wheel starts from rest, \( KE_{rot, initial} = \frac{1}{2} I \omega_i^2 = 0 \).
Therefore, the equation simplifies to:
\[ W = \frac{1}{2} I \omega_f^2 \]Substituting the calculated values for \( W \) and \( I \):
\[ 20 \, \text{J} = \frac{1}{2} (0.1 \, \text{kg} \cdot \text{m}^2) \omega_f^2 \]Final Computation & Result:
Now, we rearrange the equation to solve for the final angular velocity, \( \omega_f \).
\[ \omega_f^2 = \frac{2 \times W}{I} = \frac{2 \times 20}{0.1} \] \[ \omega_f^2 = \frac{40}{0.1} = 400 \]Taking the square root of both sides:
\[ \omega_f = \sqrt{400} = 20 \, \text{rad/s} \]The angular velocity of the wheel after the cord is unwound by 1 m will be 20 rad/s.
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