A wheel of radius $ 0.2 \, \text{m} $ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $ 10 \, \text{N} $. The established torque produces an angular acceleration of $ 2 \, \text{rad/s}^2 $. Moment of inertia of the wheel is............. kg m².
A wheel of radius $ 0.2 \, \text{m} $ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $ 10 \, \text{N} $. The established torque produces an angular acceleration of $ 2 \, \text{rad/s}^2 $. Moment of inertia of the wheel is............. kg m².
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Correct Answer: 1
Approach Solution - 1
To determine the moment of inertia (I) of the wheel, we start by using the relationship between torque (τ), moment of inertia (I), and angular acceleration (α). The equation is:
\( \tau = I \cdot \alpha \)
Given the force applied (F) and the radius (r) of the wheel, the torque can be calculated as:
\( \tau = F \cdot r \)
Substituting the given values, we have:
\( \tau = 10 \, \text{N} \times 0.2 \, \text{m} = 2 \, \text{Nm} \)
Next, use this torque in the relation \( \tau = I \cdot \alpha \) to solve for the moment of inertia (I):
\( I = \frac{\tau}{\alpha} = \frac{2 \, \text{Nm}}{2 \, \text{rad/s}^2} = 1 \, \text{kg m}^2 \)
Thus, the moment of inertia of the wheel is \( 1 \, \text{kg m}^2 \). This result fits exactly within the given range (1, 1).
Approach Solution -2
We are given:
- Radius of the wheel, \( r = 0.2 \, \text{m} \)
- Force applied, \( F = 10 \, \text{N} \)
- Angular acceleration, \( \alpha = 2 \, \text{rad/s}^2 \)
We need to find the moment of inertia (I) of the wheel.
Concept Used:
The torque (\( \tau \)) produced by a force \( F \) applied tangentially at radius \( r \) is:
\[ \tau = F \times r \]
Also, the relation between torque, moment of inertia, and angular acceleration is:
\[ \tau = I \alpha \]
Therefore, the moment of inertia can be calculated as:
\[ I = \frac{\tau}{\alpha} = \frac{F r}{\alpha} \]
Step-by-Step Solution:
Step 1: Calculate the torque produced by the applied force.
\[ \tau = F \times r = 10 \times 0.2 = 2 \, \text{Nm} \]
Step 2: Use the relation between torque and angular acceleration to find \( I \):
\[ I = \frac{\tau}{\alpha} = \frac{2}{2} = 1 \, \text{kgm}^2 \]
Final Computation & Result:
The moment of inertia of the wheel is:
\[ \boxed{I = 1 \, \text{kgm}^2} \]
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