Question

A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinder and its free end is attached to a block of mass 53.0 kg that rests on a platform. The cylinder has a mass of 275 kg and a radius of 0.360 m. Assume that the cylinder can rotate about its axis without any friction and the rope is of negligible mass. The platform is suddenly removed from under the block. The block falls down toward the ground and as it does so, it causes the rope to unwind and the cylinder to rotate.
(a)What is the angular acceleration of the cylinder? rad/s2

(b) How many revolutions does the cylinder make in 5 s? rev

(c)How much of the rope unwinds in this time interval? m

Answer 1

Answer:

(a)10.5 rad/s2

(b) 20.9 rev

(c) 47.27 m

Explanation:

As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law

T = mg = 53*9.81 = 519.93  N

Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is

To = TR = 519.93  * 0.36 = 187.17 Nm

This solid cylinder would have a moment of inertia around it's rotating axis of:

$I = \frac{mR^2}{2} = \frac{275 * 0.36^2}{2} = 17.82kgm^2$

(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder

$\alpha = \frac{To}{I} = \frac{187.17}{17.82} = 10.5 rad/s^2$

(b) With such constant angular acceleration, the angle it would make after 5s is

$\theta = \frac{\alphat^2}{2} = \frac{10.5*5^2}{2} = 131.3 rad$

Since each revolution equals to $2\pi rad$ of angle, we can calculate the number of revolution it makes

$\frac{\theta}{2\pi} = \frac{131.3}{6.28} \approx 20.9 rev$

(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad

$\theta R = 131.3*0.36 = 47.27 m$