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

How can I solve this question: A parabolic bowl with its bottom at origin has the shape y =(x^2) /20. What is the maximum height where a bowl of mass m can be placed without slipping (x and y in metres, μ=0.5)?
3 years ago

Vikas TU
10573 Points

Friction force would act tangentially obvious.
Hence, f = umg
Now, tanthetha = 2x/20 => x/10
sinthetha = x/root(x^2 + 100)
Hence,
mgsin(thetha) = f = umg
0.5 = x/root(x^2 + 100)
x^2/(x^2 + 100) = ¼
4x^2 = x^2 + 100
3x^2 = 100
x = 10/root(3) meter.
3 years ago
Amy
13 Points

dy/dx=2x/10 i.e. x/5 which is equal to tan theta
Also umg cos theta = mg sin theta
I.e. u = tan theta
Hence 1/2= X/5
Therefore value of X=2.5 m
Putting value of X in equation y=x^2/10
We get. Y= 2.5*2.5/10
Y= 0.625 m or 62.5 cm
7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

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

### Course Features

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