A small block of mass 100 g is tied to a spring of spring constant 7.5 N/m and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s about point A, then tension in the spring is-
- 0.75N
- 1.5N
- 0.25N
- 0.50N
The Correct Option is A
Approach Solution - 1
𝑘𝑥 = 𝑚𝜔^2 𝑟
⟹ 𝑘𝑥 = 0.1 × 25 × (0.2 + 𝑥)
⟹ 7.5𝑥 = 2.5(0.2 + 𝑥) ⟹ 3𝑥 = 0.2 + 𝑥
⟹ 2𝑥 = 0.2 ⟹ 𝑥 = 0.1𝑚
Now, tension in the spring = 𝑘𝑥 = 7.5 × 0.1 𝑁 = 0.75 N
Approach Solution -2
Centripetal Force and Spring Problem
Step 1: Centripetal Force
The block moves in a circular path due to the tension in the spring. This tension provides the necessary centripetal force for circular motion. The centripetal force is given by:
\( F_c = m \omega^2 r \)
where \( m \) is the mass, \( \omega \) is the angular velocity, and \( r \) is the radius of the circular path.
Step 2: Spring Force
The tension in the spring is given by Hooke's Law:
\( T = kx \)
where \( k \) is the spring constant and \( x \) is the extension of the spring.
Step 3: Relate Centripetal Force and Spring Force
The radius of the circular path is the original length of the spring (\( l \)) plus the extension (\( x \)): \( r = l + x \). The tension in the spring provides the centripetal force, so:
\( kx = m \omega^2 (l + x) \)
Step 4: Substitute Given Values and Solve for x
Given: \( m = 100 \text{ g} = 0.1 \text{ kg} \), \( k = 7.5 \text{ N/m} \), \( l = 20 \text{ cm} = 0.2 \text{ m} \), and \( \omega = 5 \text{ rad/s} \). Substituting these values:
\( 7.5x = 0.1 \times (5)^2 \times (0.2 + x) \)
\( 7.5x = 0.1 \times 25 \times (0.2 + x) \)
\( 7.5x = 2.5 (0.2 + x) \)
\( 7.5x = 0.5 + 2.5x \)
\( 5x = 0.5 \)
\( x = 0.1 \text{ m} \)
Step 5: Calculate Tension
Now that we know the extension \( x \), we can calculate the tension in the spring:
\( T = kx = 7.5 \times 0.1 = 0.75 \text{ N} \)
Conclusion:
The tension in the spring is 0.75 N (Option 1).
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