Answer :
To solve this problem, let's break it into manageable steps and understand the mechanics involved in circular motion.
1. Understanding the Question:
- A particle of mass [tex]\(m = 1 \, \text{kg}\)[/tex] is moving in a circular path of radius [tex]\(r = 1 \, \text{m}\)[/tex].
- The maximum tension the string can withstand (which is the maximum centripetal force) is [tex]\(T = 9.8 \, \text{N}\)[/tex].
- We are asked to find the change in speed and the change in velocity of the particle as it moves 60 degrees (Ï€/3 radians) around the circle.
2. Calculating Angular Velocity ([tex]\(\omega\)[/tex]):
- The tension [tex]\(T\)[/tex] provides the centripetal force needed for circular motion:
[tex]\[ T = m r \omega^2 \][/tex]
- Solving for angular velocity ([tex]\(\omega\)[/tex]):
[tex]\[ \omega = \sqrt{\frac{T}{m r}} \][/tex]
Plugging in the values:
[tex]\[ \omega = \sqrt{\frac{9.8}{1 \cdot 1}} = \sqrt{9.8} \approx 3.1305 \, \text{rad/s} \][/tex]
3. Calculating Initial Speed (v):
- The linear speed [tex]\(v\)[/tex] is related to angular velocity by:
[tex]\[ v = r \omega \][/tex]
Plugging in the values:
[tex]\[ v = 1 \cdot 3.1305 \approx 3.1305 \, \text{m/s} \][/tex]
4. Change in Speed:
- In uniform circular motion, the speed [tex]\(v\)[/tex] remains constant. Therefore, the change in speed is:
[tex]\[ \Delta v = 0 \][/tex]
5. Change in Velocity:
- Although the speed remains constant, the direction of the velocity vector changes as the particle moves along the circular path.
- The change in velocity over an angle [tex]\(\theta\)[/tex] (in radians) can be calculated using:
[tex]\[ \Delta \text{velocity} = v \cdot 2 \cdot \sin\left(\frac{\theta}{2}\right) \][/tex]
- For [tex]\(\theta = 60^\circ = \frac{\pi}{3}\)[/tex]:
[tex]\[ \Delta \text{velocity} = 3.1305 \cdot 2 \cdot \sin\left(\frac{\pi}{6}\right) = 3.1305 \cdot 2 \cdot 0.5 \approx 3.1305 \, \text{m/s} \][/tex]
Summary:
- The change in speed is 0.
- The change in velocity is approximately 3.1305 m/s.
Therefore, the correct option is:
A) [tex]\(0, v 2v\)[/tex]
1. Understanding the Question:
- A particle of mass [tex]\(m = 1 \, \text{kg}\)[/tex] is moving in a circular path of radius [tex]\(r = 1 \, \text{m}\)[/tex].
- The maximum tension the string can withstand (which is the maximum centripetal force) is [tex]\(T = 9.8 \, \text{N}\)[/tex].
- We are asked to find the change in speed and the change in velocity of the particle as it moves 60 degrees (Ï€/3 radians) around the circle.
2. Calculating Angular Velocity ([tex]\(\omega\)[/tex]):
- The tension [tex]\(T\)[/tex] provides the centripetal force needed for circular motion:
[tex]\[ T = m r \omega^2 \][/tex]
- Solving for angular velocity ([tex]\(\omega\)[/tex]):
[tex]\[ \omega = \sqrt{\frac{T}{m r}} \][/tex]
Plugging in the values:
[tex]\[ \omega = \sqrt{\frac{9.8}{1 \cdot 1}} = \sqrt{9.8} \approx 3.1305 \, \text{rad/s} \][/tex]
3. Calculating Initial Speed (v):
- The linear speed [tex]\(v\)[/tex] is related to angular velocity by:
[tex]\[ v = r \omega \][/tex]
Plugging in the values:
[tex]\[ v = 1 \cdot 3.1305 \approx 3.1305 \, \text{m/s} \][/tex]
4. Change in Speed:
- In uniform circular motion, the speed [tex]\(v\)[/tex] remains constant. Therefore, the change in speed is:
[tex]\[ \Delta v = 0 \][/tex]
5. Change in Velocity:
- Although the speed remains constant, the direction of the velocity vector changes as the particle moves along the circular path.
- The change in velocity over an angle [tex]\(\theta\)[/tex] (in radians) can be calculated using:
[tex]\[ \Delta \text{velocity} = v \cdot 2 \cdot \sin\left(\frac{\theta}{2}\right) \][/tex]
- For [tex]\(\theta = 60^\circ = \frac{\pi}{3}\)[/tex]:
[tex]\[ \Delta \text{velocity} = 3.1305 \cdot 2 \cdot \sin\left(\frac{\pi}{6}\right) = 3.1305 \cdot 2 \cdot 0.5 \approx 3.1305 \, \text{m/s} \][/tex]
Summary:
- The change in speed is 0.
- The change in velocity is approximately 3.1305 m/s.
Therefore, the correct option is:
A) [tex]\(0, v 2v\)[/tex]
