Question

A motorcycle, which has an initial linear speed of 5.0 m/s, decelerates to a speed of 3.5 m/s in 4.5 s. Each wheel has a radius of 0.60 m and is rotating in a counterclockwise (positive) directions. What is (a) the constant angular acceleration (in rad/s2) and (b) the angular displacement (in rad) of each wheel?

Answer 1

(a) The angular acceleration is - 0.56 rad/s^2.

(b) The angular displacement is 31.6 rad.

Step-by-step explanation:

initial velocity, u = 5 m/s

final velocity, v = 3.5 m/s

radius, r = 0.6 m

time, t = 4.5 s

initial angular velocity, wo = u/r = 5/0.6 = 8.33 rad/s

final angular velocity, w = v/r = 3.5 / 0.6 = 5.83 rad/s

(a) Use the first equation of motion to fine the angular acceleration.

`w = w_o + \alpha t \\\\5.83 = 8.33 + \alpha * 4.5\\\\\alpha = - 0.56 rad/s^2`

(b) Use third equation of motion to find the angular displacement

`w^2 = w_0^2 + 2\alpha \theta \\\\5.83^2 =8.833^2 - 2 * 0.56* \theta \\\\\theta =31.6 rad`

Answer 2

Step-by-step explanation:

Given that,

A motorcycle, which has an initial linear speed of 5.0 m/s, decelerates to a speed of 3.5 m/s in 4.5 s. Each wheel has a radius of 0.60 m and is rotating in a counterclockwise (positive) directions.

Angular acceleration,
`\alpha =(\omega_2-\omega_1)/(t)`

`\alpha =((v_2)/(r)-(v_1)/(r))/(t)`

Put all the values,

`\alpha =((3.5)/(0.6)-(5)/(0.6))/(4.5)\\\\=-0.56\ rad/s^2`

Angular displacement,

`\theta=(\omega_2^2-\omega_1^2)/(2\alpha)\\\\\theta=(((v_2)/(r))^2-((v_1)/(r))^2)/(2\alpha)\\\\\theta=(((3.5)/(0.6))^2-((5)/(0.6))^2)/(2* (-0.56))\\\\=31.62\ rad`

Hence