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A thin uniform circular ring is rolling down an inclined plane of inclination 300 without slipping. Its linear acceleration along the inclined plane will be
- a)g/2
- b)g/3
- c)g/4
- d)2g/3
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A thin uniform circular ring is rolling down an inclined plane of incl...
Acceleration on an inclined plane
a= g sinθ/ (1 + I/MR^2)
for circular rings: I= MR^2
so by putting the value in above equation we get
a= g sinθ/ 2
a = g sin30/2
a= g/4
Most Upvoted Answer
A thin uniform circular ring is rolling down an inclined plane of incl...
Consider a thin uniform circular ring rolling down in an
inclined plane without slipping. Compute the linear
acceleration along the inclined plane if the angle of
inclination is 45˚
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A thin uniform circular ring is rolling down an inclined plane of incl...
To find the linear acceleration of a thin uniform circular ring rolling down an inclined plane, we can use the principles of rotational motion and Newton's second law of motion.
1. Identify the given information:
- The inclined plane has an inclination of 30 degrees.
- The ring is rolling down the inclined plane without slipping.
2. Determine the forces acting on the ring:
- The weight of the ring acts vertically downwards and can be resolved into two components:
- The component parallel to the inclined plane (mg*sin(30°)) causes the ring to accelerate down the plane.
- The component perpendicular to the inclined plane (mg*cos(30°)) does not contribute to the acceleration.
- The normal force exerted by the inclined plane on the ring acts perpendicular to the plane and balances the perpendicular component of the weight.
3. Apply Newton's second law of motion:
- The net force acting on the ring is the component of the weight parallel to the inclined plane (mg*sin(30°)).
- According to Newton's second law, F = ma, where F is the net force and a is the linear acceleration.
- Therefore, mg*sin(30°) = ma.
4. Solve for the linear acceleration:
- Divide both sides of the equation by the mass of the ring (m) to get the linear acceleration (a) alone.
- a = g*sin(30°).
5. Calculate the value of the linear acceleration:
- The value of g is approximately 9.81 m/s^2.
- sin(30°) is 0.5.
- Therefore, a = 9.81 m/s^2 * 0.5 = 4.905 m/s^2.
6. Compare the calculated value with the given options:
- The option 'C' states that the linear acceleration is g/4, which is equivalent to 9.81 m/s^2 / 4 = 2.4525 m/s^2.
- The calculated value of the linear acceleration (4.905 m/s^2) is approximately twice the value given in option 'C'.
- Since the calculated value is larger than the value given in option 'C', it is not the correct answer.
Therefore, the correct answer is not option 'C'.
1. Identify the given information:
- The inclined plane has an inclination of 30 degrees.
- The ring is rolling down the inclined plane without slipping.
2. Determine the forces acting on the ring:
- The weight of the ring acts vertically downwards and can be resolved into two components:
- The component parallel to the inclined plane (mg*sin(30°)) causes the ring to accelerate down the plane.
- The component perpendicular to the inclined plane (mg*cos(30°)) does not contribute to the acceleration.
- The normal force exerted by the inclined plane on the ring acts perpendicular to the plane and balances the perpendicular component of the weight.
3. Apply Newton's second law of motion:
- The net force acting on the ring is the component of the weight parallel to the inclined plane (mg*sin(30°)).
- According to Newton's second law, F = ma, where F is the net force and a is the linear acceleration.
- Therefore, mg*sin(30°) = ma.
4. Solve for the linear acceleration:
- Divide both sides of the equation by the mass of the ring (m) to get the linear acceleration (a) alone.
- a = g*sin(30°).
5. Calculate the value of the linear acceleration:
- The value of g is approximately 9.81 m/s^2.
- sin(30°) is 0.5.
- Therefore, a = 9.81 m/s^2 * 0.5 = 4.905 m/s^2.
6. Compare the calculated value with the given options:
- The option 'C' states that the linear acceleration is g/4, which is equivalent to 9.81 m/s^2 / 4 = 2.4525 m/s^2.
- The calculated value of the linear acceleration (4.905 m/s^2) is approximately twice the value given in option 'C'.
- Since the calculated value is larger than the value given in option 'C', it is not the correct answer.
Therefore, the correct answer is not option 'C'.
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A thin uniform circular ring is rolling down an inclined plane of inclination 300 without slipping. Its linear acceleration along the inclined plane will bea)g/2b)g/3c)g/4d)2g/3Correct answer is option 'C'. Can you explain this answer?
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