Question

At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.2 m/s2 .\ Part A
Calculate the wheel's constant angular acceleration.

Part B

Calculate the angular velocity at t = 3.40s .

Part C

Calculate the angular velocity at t=0.

Part D

Through what angle did the wheel turn between t=0 and t = 3.40s ?

Part E

Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81m/s2 ?

Answer 1

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s