Home
Class 11
PHYSICS
A particle starts rotating from rest acc...

A particle starts rotating from rest according to the formula `theta = (3 t^(3)//20) - (t^(2)//3)` .Find the angular velocity and the acceleration at the end of 5s.

Text Solution

Verified by Experts

`omega = d theta//dt = 9t^(2)//20-2t//3`, at `t=5s, omega = 7.92 "rad"s^(-1), alpha = domega//dt = 18t//20 - 2//3`, at t=5s, `alpha = 3.83 "rads"^(-1)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CIRCULAR MOTION

    ICSE|Exercise MODULE 2 (FROM CENTRE OF MASS)|8 Videos

  • CIRCULAR MOTION

    ICSE|Exercise MODULE 2 (FROM MOMENT OF INERTIA, TORQUE)|22 Videos

  • CIRCULAR MOTION

    ICSE|Exercise MODULE 2 (VERY SHORT ANSWER QUESTIONS)|14 Videos

  • COMPETITION CARE UNIT

    ICSE|Exercise OBJECTIVE QUESTIONS FROM PREVIOUS IAS EXAMINATIONS |50 Videos

Similar Questions

Explore conceptually related problems

A particle starts rotating from rest according to the formula theta = (t^(2)//64) - (t//8) " where " theta is the angle in radian and t in s . Find the angular velocity and angular acceleration at the end of 4 s .

A particle starts from rest and its angular displacement (in rad) is given theta=(t^2)/(20)+t/5 . Calculate the angular velocity at the end of t=4s .

The position x of a particle varies with time t according to the relation x=t^3+3t^2+2t . Find the velocity and acceleration as functions of time.

The angular displacement of a body is given by theta = 2t^(2) + 5t -3 . Find the value of the angular velocity and angular acceleration when t = 2s.

The angular position of a point over a rotating flywheel is changing according to the relation, theta = (2t^3 - 3t^2 - 4t - 5) radian. The angular acceleration of the flywheel at time, t = 1 s is

Check the correctness of the formula S = ut + 1/3 "at"^2 where S is the distance , u is velocity , a is acceleration and t is time.

A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation v = (5t^(2) - 3t + 2)m//s . Now give the answer of following questions : Acceleration of particle at the end of 3^(rd) second of motion is :

A particle moves along a staight line such that its displacement at any time t is given by s=t^3-6t^2+3t+4m . Find the velocity when the acceleration is 0.

The acceleration a of a particle starting from rest varies with time according to relation, a=alphat+beta . Find the velocity of the particle at time instant t. Strategy : a=(dv)/(dt)

The velocity of a particle is given by v=(2t^(2)-3t+10)ms^(-1) . Find the instantaneous acceleration at t = 5 s.

Home
Profile