 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
When a uniform solid sphere mass m rolls from rest down an inclined plane such that its centre of mass descends through a height h, its translational kinetic energy will be

```
3 years ago

```							5/7mg is the answer if a solid sphere mass m rolls from rest down an inclined plane its centre if mass desend height h its tramsitional kinetic energy will be
```
one year ago
```							dear studentfor a sphere rolling down an inclined plane (without slipping) the following condition existspotential energy at top = translational kinetic energy at bottom + rotational kinetic energy at bottomso,PE = KEt + KErormgh = (1/2)mv^2 + (1/2)Iw^2hereI = moment of inertia = (2/5)MR^2and v = translational velocity = R.wthus, we havemgh = (1/2)M.R^2w^2 + (1/2).(2/5).MR^2.w^2orgh = (7/10).)R^2w^2or the angular velocity becomesw2 = (10/7).[gh/R^2]we can now use this value of w2 to calculate the translational energyEt = (1/2)mv^2 + (1/2)Iw^2 = (1/2)M.R^2w^2 orEt = (1/2)MR^2.[((10/7).[gh/R^2])] orEt = (5/7)M.[gh]orE t = (5/7)Mghandthe rotational kinetic energy will beEr = mgh - (5/7)MghorE r = (2/7)Mgh
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions