The non-constant angular acceleration of a wheel is given by a(t)=At. A=6 radian

Question

The non-constant angular acceleration of a wheel is given by a(t)=At. A=6 radians/s^3. At t=1.00second, the wheels angular velocity is 17 radians/s and its angular position is 700 radians.

a) what is the angular velocity at t=2.00 seconds
b)what is the average angular acceleration from t=1.00 second to t=2.00 seconds.
c) calculate the instantaneous angular acceleration at t=1.50 seconds.
d) find an expression for the angular velocity at any time t.
e) what is the angular position at t=2.00 seconds.
f) find an expression for the angular position at any time t.

Explanation / Answer

(a) (t)=At where A=6rad/s3 t=1.00s: =17rad/s and =700rad

(t)dt=6tdt => (t)=3t2+C and (1)=17 so 17=3(1)2+C, C=14. So (t)=3t2+14

(2)=3(2)2+14=26 so at t=2s is 26rad/s

(b)avg=/t=(f-o)/(tf-to)=(26-17)/(2-1)=(9/1) so avg=9rad/s2

(c) (t)=6t so (1.50)=6(1.50)=9 so at t=1.50s =9rad/s2

(d) this was found in part (a): (t)=3t2+14

(e) (t)dt=(3t2+14)dt => (t)=t3+14t+C and (1)=700 so 700=(1)3+14(1)+C, C=685 So (t)=t3+14t+685

now t=2 so (2)=(2)3+14(2)+685=721 so at t=2s is 721rads

(f) this was found in part (e): (t)=t3+14t+685

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