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A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to0.200rad/s2, what is its angular velocity at t = 2.50 s ? (b) Through what angle has the wheel turned between t = 0 and t=2.50s ?

Short Answer

Expert verified

(a) The required angular speed is 2 rad/s .

(b) The required angle is 4.375 rad .

Step by step solution

01

Identification of given data

The angular speed isÓ¬0z=1.5rad/s .

The angular acceleration isαz=0.200rad/s2 .

The time ist=2.50s .

02

Concept/Significance of rotation of body with constant angular acceleration

Angular velocity at time t of a rigid body with constant angular acceleration is given by,

Ӭz=Ӭ0z+αzt........(1)

Here,αz is the angular acceleration,Ӭ0z is the angular velocity of body at time 0, is the time.

Angular position at time t of a rigid body with constant angular acceleration is given by,

θ=Ӭ0zt+12αzt2......(2)

03

Determine the angular velocity at t=2.50s(a)

SubstituteӬ0z=1.5rad/s,t=2.5s,andαz=0.200rad/s2 in equation (1).

Ó¬z=1.5rad/s+0.200rad/s22.5s=1.5rad/s+0.5rad/s=2rad/s

Therefore, the required angular speed is 2 rad/s .

04

Determine the time angle turned by wheel between t = 0 and t = 2.50 s(b)

Substitute Ӭ0z=1.5rad/s,t=2.5sandαz=0.200rad/s2 in equation (2).

θ=1.5rad/s2.5s+120.200rad/s22.5s2=3.75rad+0.625rad=4.375rad

Therefore, the required angle is 4.375 rad .

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