Question

A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string. When the system is released from rest, you determine that the stone reaches a speed of 3.40 m/s after having fallen 2.40 m. What is the mass of the cylinder?

Answer 1

Answer:18.5 kg

Explanation:

Given

diameter $d=75 cm$

mass of stone $m=3 kg$

velocity of stone $v=3.4 m/s$

height fallen $h=2.40 m$

using

$v^2-u^2=2as$

$(3.4)^2=2\times a\times 2.4$

$a=2.4 m/s^2$

Tension of string will Provide Torque to cylinder

$T\times r=I\times \alpha$

where $I=moment\ of\ inertia$

$\alpha =angular\ acceleration$

$T=\frac{Mr^2}{2}\times \frac{a}{r}$

$T=\frac{Ma}{2}$

and $mg-T=ma$

$mg-ma=T$

Put value of T

$mg-ma=\frac{Ma}{2}$

$mg=a(m+\frac{M}{2})$

$3\times 9.8=2.4\cdot (m+\frac{M}{2})$

$12.25=3+\frac{M}{2}$

$9.25=\frac{M}{2}$

$M=18.5 kg$

Answer 2

Since a light string is wrapped around the outer rim of a solid uniform cylinder, the mass of the cylinder is equal to 18.4 kilograms.

Given the following data:

• Diameter = 75.0 cm

• Mass of stone = 3 kg

• Initial velocity = 0 m/s (since the system was released from rest)

• Final velocity = 3.40 m/s

• Distance = 2.40 meters.

To calculate the mass of the cylinder:

First of all, we would determine the acceleration of the stone by using the the third equation of motion;

$V^2 = U^2 + 2aS$

Where:

• V is the final speed.

• U is the initial speed.

• a is the acceleration.

• S is the distance covered.

Substituting the given parameters into the formula, we have;

$3.4^2 = 0^2 + 2a(2.40)\\\\11.56 = 0 + 4.8a\\\\11.56 = 4.8a\\\\a = \frac{11.56}{4.8}$

Acceleration, a = 2.41 $m/s^2$

Next, we would solve for the torque by using the formula:

$T = m(g - a)\\\\T = 3(9.8 - 2.41)\\\\T = 3 \times 7.39$

Torque, T = 22.17 Newton.

In rotational motion, torque is given by the formula:

$T = I\alpha \\\\Tr = \frac{Mr^2}{2} \times \frac{a}{r} \\\\Tr = \frac{Mra}{2}\\\\2Tr = Mra\\\\M = \frac{2T}{a}\\\\M = \frac{2\times 22.17}{2.41}\\\\M = \frac{44.34}{2.41}$

Mass = 18.4 kilograms

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