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A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will be
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A disc of radius b and mass m rolls down an inclined plane of vertical...
The difference in the potential of the body when it rolls down through a vertical height h, is mgh.
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½ mv2 + ¼ mv2
Thus we get v = √gh/3
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½ mv2 + ¼ mv2
Thus we get v = √gh/3
Most Upvoted Answer
A disc of radius b and mass m rolls down an inclined plane of vertical...
The difference in the potential of the body when it rolls down through a vertical height h, is mgh.
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½ mv2 + ¼ mv2
Thus we get v = √gh/3
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½ mv2 + ¼ mv2
Thus we get v = √gh/3
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A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will bea)b)c)d)Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will bea)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will bea)b)c)d)Correct answer is option 'A'. Can you explain this answer?.
A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will bea)b)c)d)Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will bea)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will bea)b)c)d)Correct answer is option 'A'. Can you explain this answer?.
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