Question

A wheel of radios 0.5m initialy at rest with a constant angular acceleration and attains a linear speed of 30 m/s in 5 sec. calculate

a) the angular acceleration?
b) the angular velocit's after 4 sec?
c) the number of revolution at a time of 5sec?​

Answer 1

a.) 12
`(rad)/(s^2)`

b.) 48
`(rad)/(s)`

c.) 23.87 rev

Step-by-step explanation:

First, convert linear velocity to angular velocity. The angular velocity can be calculated using the relationship v=rω, where v is the linear velocity, r is the radius, and ω is the angular velocity. Rearrange the equation to solve for ω:

`\omega=(v)/(r)`

For this problem, let v=30
`(m)/(s)` and r=0.5m. So,

`\omega=(30(m)/(s))/(0.5m)\\\omega=60(rad)/(s)`

Next, to solve for angular acceleration use rotational motion equation #1,
`\omega_f=\omega_i+\alpha t`. For this problem, let

`\omega_i=0.0(rad)/(s)\\\omega_f=60(rad)/(s)\\t=5s`

So,

`60(rad)/(s)=0.0(rad)/(s)+\alpha(5s)\\((1)/(5s))60(rad)/(s)=\alpha(5s)((1)/(5s))\\12(rad)/(s^2)=\alpha`

Next, use the angular acceleration and the time given in part b in rotational motion equation #1 to solve for the angular velocity at t=4s:

`\omega_f=\omega_i+\alpha t\\\omega_f=0.0(rad)/(s)+(12(rad)/(s^2))(4s)\\\omega_f=48(rad)/(s)`

To find the number of revolutions, use rotational motion equation #3,
`\theta=v_it+(1)/(2)\alpha t^2`, to find the angular displacement at t=5s. So,

`\theta=0.0(m)/(s)(5s)+(1)/(2)(12.0(rad)/(s^2))(5s)^2\\\theta=0.0m+150\ rad\\\theta=150\ rad`

To convert the angular displacement in radians to revolutions, recall that one revolution equals 2π radians. So,

`150\ rad((1\ rev)/(2\pi\ rad))=(75)/(\pi)\ rev\approx23.87\ rev`