Thank you for registering.

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

Please check your email for login details.
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

A stone of mass m is whirled in a vertical circle with a string of length l. At an instant, the stone is at its lowest position and has velocity u. Find the change in magnitude of its velocity when the string is horizontal.

A stone of mass m is whirled in a vertical circle with a string of length l. At an instant, the stone is at its lowest position and has velocity u. Find the change in magnitude of its velocity when the string is horizontal.

Grade:12

3 Answers

Naveen Kumar
askIITians Faculty 60 Points
6 years ago
Initial kinetic energy of block (at bottom)=(1/2)*m*u2
let
At the time when the string is horizontal,the velocity is V(say) and so the kinetic energy=(1/2)*m*V2
during the process,gain iun potential energy=mg*l
Applying energy conservation,
(1/2)*m*u2 = (1/2)*m*V2........+.........mg*l
V=sq.root(u2-2g*l)
and hence, u-V=u-sq.root(u2-2g*l)
Rupali Sehgal
21 Points
6 years ago
sir from the options given, answer is  sq root 2(u2-gl). How does that come?
Mallikarjun Maram
50 Points
6 years ago
u and v are perpendicular to each other.  Hence magnitude of their difference will be equal to sqrt(u2 + u2 – 2gl) = sqrt(2(u2 – gl))
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free