#### Thank you for registering.

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

Please check your email for login details.

Click to Chat

1800-1023-196

+91 7353221155

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 solid body rotates about a stationary axis according to the law theta=at-bt^3,where a=6 rad/s and b=2 rad/s^3.Find the mean values of the angular velocity and acceleration over the interval between t=0 and the time,when the body comes to rest.

## 1 Answers

10 years ago

@ = at - bt3               ...............1

d@/dt = W = a-3bt2             ...............2

dW/dt = alfa = -6bt           ............3

body comes to rest when angular velocity  becomes 0

a - 3bt2 = 0

t = (a/3b)1/2

mean value of angular velocity = (@t - @0)/t

@0=0

@t= a(a/3b)1/2 - b(a/3b)3/2

= a(a/3b)1/2[1-1/3]

=(2a/3)(a/3b)1/2

mean value of angular velocity = 2a/3(a/3b)1/2/t = 2a/3 = 4rad/sec

now mean value of angular accleration = change in angular velocity/t = (Wt-W0)/t

at t=0  W=a , at T=t  Wt = 0

mean value = a/t = (3ab)1/2

=(36)1/2=6rad/sec2

## ASK QUESTION

Get your questions answered by the expert for free