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
Menu
kriti Grade: 11
        
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)?
one year ago

Answers : (1)

Vikas TU
6872 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.
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details