Home
Class 12
PHYSICS
A swing moving in a children's garden is...

A swing moving in a children's garden is observed to move with an angular velocity given by, `omega-a(t^2)hat(i)+b(e^-1)hat(j)`. What will be the angle between angular acceleration and angular velocity at t=1 s given that a=b=1 unit?

A

`20^@`

B

`36^@`

C

`15^@`

D

`9^@`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • CIRCULAR MOTION

    TARGET PUBLICATION|Exercise COMPETITIVE THINKING|102 Videos

  • ATOMS, MOLECULES AND NUCLEI

    TARGET PUBLICATION|Exercise Evaluation test|19 Videos

  • CURRENT ELECTRICITY

    TARGET PUBLICATION|Exercise Evaluation Test|6 Videos

Similar Questions

Explore conceptually related problems

What is the angular acceleration of a particle moving with constant angular velocity ?

The angular velocity of a particle is given by omega=1.5t-3t^(@)+2 , Find the time when its angular acceleration becomes zero.

The angular velocity of a particle is given by the equation omega =2t^(2)+5rads^(-1) . The instantaneous angular acceleration at t=4 s is

A particle moving along a circule path of radius 'r' with uniform angular velocity omega . Its angular acceleration is

If angular velocity of a disc depends an angle rotated theta as omega=theta^(2)+2theta , then its angular acceleration alpha at theta=1 rad is :

the angular velocity omega of a particle varies with time t as omega = 5t^2 + 25 rad/s . the angular acceleration of the particle at t=1 s is

A solid body rotates with angular velocity vecomega=3thati+2t^(2) hatj rad//s . Find (a) the magnitude of angular velocity and angular acceleration at time t=1 s and (b) the angle between the vectors of the angular velocity and the angular acceleration at that moment.

The rod of length l=1 m rotates with an angular velocity omega=2 rads^(-1) an the point P moves with velocity v=1ms^(-1) and acceleration a=1ms^(-2) . Find the velocity and acceleration of Q .

The angular position of a swinging door is described by theta=5+10t+20t^(2) . Determine the angular position, angular speed and angular acceleration at t=2 s .

The angular position of a point on the rim of a rotating wheel is given by theta=4t^(3)-2t^(2)+5t+3 rad. Find (a) the angular velocity at t=1 s , (b) the angular acceleration at t=2 s . (c ) the average angular velocity in time interval t=0 to t=2 s and (d) the average angular acceleration in time interval t=1 to t=3 s .