- #1

I_Try_Math

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- Homework Statement
- A cord is wrapped around the rim of a solid cylinder of radius 0.25 m, and a constant force of 40 N is exerted on the cord shown, as shown in the following figure. The cylinder is mounted on frictionless bearings, and its moment of inertia is ##6 kg \cdot m^2##. (a) Use the work energy theorem to calculate the angular velocity of the cylinder after 5.0 m of cord have been removed. (b) If the 40-N force is replaced by a 40-N weight, what is the angular velocity of the cylinder after 5.0 m of cord have unwound?

- Relevant Equations
- ##W_{AB} = K_B - K_A##

## a_t = r\alpha ##

## \omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta) ##

## a_t = r\alpha ##

## 9.8 \: m/s^2 = (.25 \: m)\alpha ##

## \alpha = 39.2 \: rad/s^2 ##

Since the cylinder rotates through 20.01 radians after 5 m of cord has been removed:

## \omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta) ##

## \omega_f^2 = 2(39.2 \: rad/s^2)(20.01 \: rad) ##

## \omega_f = 39.6 \: rad/s ##

Textbook answer key says ## \omega_f = 8 \: rad/s ##. Can't see where my calculation is wrong. Any help is appreciated.