At a time t = 3.50 s , a point on the rim of a wheel with a radius of 0.250 m ha

Question

At a time t = 3.50 s , a point on the rim of a wheel with a radius of 0.250 m has a tangential speed of 51.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.3 m/s2 .

A.Calculate the wheel's constant angular acceleration.

B.Calculate the angular velocity at

t = 3.50 s .

C.Calculate the angular velocity at

t=0.

D.Through what angle did the wheel turn between

t=0 and t = 3.50 s ?

E.Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal

g = 9.81 m/s2 ?

Explanation / Answer

here ,

at =- 10.3 m/s^2

a) t = 3.5 s

let the constant acceleration is a

a = at/r

a = 10.3/.250

a = - 41.2 rad/s^2

the wheel's constant angular acceleration is - 41.2 rad/s^2

B)

angular velocity is

w = velocity/radius

w = 51/.250 rad/s

w = 204 rad/s

the angular speed is 204 rad/s

C)

at t = 0 s

let the angular speed is wi

wi = w - a * t

wi = 204 - (-41.2) * 3.5

wi = 348.2 rad/s

the angular speed at t = 0 s is 348.2 rad/s

D)

let the angle is theta

Using forth equation of motion

wf^2 -wi^2 =2 *a * theta

204^2 - 348.2^2 = -2 * 41.2 * theta

theta = 966.4 radians

the angle rotated is 966.4 radians

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