Question

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is then pulled with a constant acceleration of 1.5 m/s2 until 1.3 m of string has been unwound. If the string unwinds without slipping, what is the cylinder's angular speed, in rpm, at this time?

Answer 1

The angular speed of the cylinder is 567.6 rpm.

Step-by-step explanation:

To find the angular speed of the cylinder, we can use the formula:

angular speed = linear speed / radius

In this case, the linear speed can be found by multiplying the acceleration by the elapsed time:

linear speed = acceleration * time

Plugging in the given values, we get:

linear speed = 1.5 m/s² * 1.3 m = 1.95 m/s

Now we can calculate the angular speed:

angular speed = 1.95 m/s / 0.033 m = 59.09 m/s

Finally, converting the angular speed to rpm:

angular speed (in rpm) = 59.09 m/s * 60 s / 2π rad = 567.6 rpm

Answer 2

`\omega_f=571.42\ rpm`

Step-by-step explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string,
`a=1.5\ m/s^2`

Displacement, d = 1.3 m

The angular acceleration is given by :

`\alpha =(a)/(r)`

`\alpha =(1.5)/(0.033)`

`\alpha =45.46\ rad/s^2`

The angular displacement is given by :

`\theta=(d)/(r)`

`\theta=(1.3)/(0.033)`

`\theta=39.39\ rad`

Using the third equation of rotational kinematics as :

`\omega_f^2-\omega_i^2=2\alpha \theta`

Here,
`\omega_i=0`

`\omega_f=√(2\alpha \theta)`

`\omega_f=√(2* 45.46* 39.39)`

`\omega_f=59.84\ rad/s`

Since, 1 rad/s = 9.54 rpm

So,

`\omega_f=571.42\ rpm`

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.