#### 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

# a particle is moving in a circle of radius R in such a way that at any instant the norml &tangential components of its accelkeration are equal if its speed at t=0 is v the time taken to complete the first revolution?

Chetan Mandayam Nayakar
312 Points
9 years ago

ω2r=αr, which implies that α=ω2,u=rω,where u is tangential velocity

u=v+αt,ω=(v+(alpha*t))/R,

ω=(v+ω2t)/R,now we have obtained a quadratic equation in omega.

Solve it to obtain omega as a function of 't'

from v=R*(omega),find v as a function of t,in addition,alpha=(omega)^2

u=v+(alpha*t)=v+ω2t

Solve it to obtain omega as a function of 't'

∫(v+αt)*dt (from t=0 to t=T)=2pi*R

you will get an equation in ' t',solve it to obtain the answer 'T'