Question

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 675 rev/min in 63.0 s? (d) How many revolutions does the wheel make during that 63.0 s?

Answer 1

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446
`rad/sec^2`

(d) 396.22 revolution

Step-by-step explanation:

We have given that diameter d = 1.63 m

So radius
`r=(d)/(2)=(1.63)/(2)=0.815m`

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

`\omega =(2\pi N)/(60)=(2* 3.14* 79.9)/(60)=8.362rad/sec`

(b) We know that linear speed is given by

`v=r\omega =0.815* 8.362=6.815m/sec`

(c) We have given final angular velocity
`\omega _f=675rev/min`

And
`\omega _i=79.9rev/min`

Time t = 63 sec

Angular acceleration is given by
`\alpha =(\omega _f-\omega _i)/(t)=(675-79.9)/(63)=9.446rad/sec^2`

(d) Change in angle is given by

`\Theta =(1)/(2)(\omega _i+\omega _f)t=(1)/(2)(675+79.9)* 1.05=396.22rev`