In: Physics

# A frictionless pulley, which can be modeled as a 0.90kg solid cylinder with a 0.31m radius

A frictionless pulley, which can be modeled as a 0.90kg solid cylinder with a 0.31m radius, has a rope going over it, as shown in the figure.  The tensions in the rope are 12N and 10N. What is the angular acceleration of the pulley?

## Solutions

##### Expert Solution

Concepts and reason

The concept required to solve this problem is rotational motion. Initially, calculate the torque using tensions and radius of the disc. Then, calculate moment of inertia of the pulley by applying the equation for moment of inertia of solid disc using mass and radius of the solid disc. Finally, calculate the angular acceleration using Newton's second law for rotation.

Fundamentals

In rotational motion, torque is crucial quantity to generate an angular acceleration of the object. The magnitude of torque $$\tau$$ due to tension in the rope is given by following expression. $$\tau=T r$$

Here, $$T$$ is the tension, and $$r$$ is the radius of the pulley that is the distance measured from the axis of rotation to where the tension is applied. The amount of torque needed to make an angular acceleration depends on the distribution of the mass of the object. This distribution is described by the moment of inertia. The moment of inertia can be found by breaking up the object into little fragments, multiplying the mass of each little parts by the square of the distance it is from the axis of rotation, and adding all these products up. Moment of Inertia, $$I=\sum m r^{2}$$ The moment of inertia of a solid disc about an axis passes through its centre is, $$I=\frac{1}{2} m r^{2}$$

Here $$m$$ is the mass of the disc, and $$r$$ is the radius of the disc. The torque from Newton's law of rotation is defined as the product of moment of inertia I and angular acceleration

$$\alpha$$

$$\tau=I \alpha$$

Calculate the net torque using tensions in the rope. The two tensions $$T_{1}, T_{2}$$ are pulling in opposite direction and the net torque on the pulley is, $$\tau=\left(T_{2}-T_{1}\right) r$$

Substitute $$12 \mathrm{~N}$$ for $$T_{2}, 10 \mathrm{~N}$$ for $$T_{1}$$ and $$0.31 \mathrm{~m}$$ for $$r$$

$$\begin{array}{c} \tau=(12 \mathrm{~N}-10 \mathrm{~N}) 0.31 \mathrm{~m} \\ =2 \mathrm{~N} \times 0.31 \mathrm{~m} \\ =0.62 \mathrm{~N} \cdot \mathrm{m} \end{array}$$

The torque due to $$T_{1}$$ is clockwise and therefore taken to be in the negative direction and the torque due to $$T_{2}$$ is in anticlockwise direction and so it is taken as positive direction. So, the difference in tension along with the radius of the pulley are used to calculate the net torque of the pulley.

Moment of inertia of solid cylinder, $$I=\frac{1}{2} m r^{2}$$

Here, $$m$$ is the mass of the solid cylinder and $$r$$ is the radius of the solid cylinder. Substitute $$0.90 \mathrm{~kg}$$ for $$m$$ and $$0.31 \mathrm{~m}$$ for $$r$$ in the equation of moment of inertia,

$$\begin{array}{c} I=\frac{1}{2} m r^{2} \\ =\frac{1}{2} \times 0.90 \mathrm{~kg} \times(0.31 \mathrm{~m})^{2} \\ =0.043245 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{array}$$

Assume the pulley as a solid disc and the equation for moment of inertia of solid cylinder is used to found moment of inertia of the pulley.

The torque from Newton's law of rotation is defined as the product of moment of inertia I and angular acceleration

$$\alpha$$

$$\tau=I \alpha$$

Rearrange the above equation for angular acceleration, $$\alpha=\frac{\tau}{I}$$

Substitute0.62N.mfor $$\tau$$ and $$0.043245 \mathrm{~kg} \cdot \mathrm{m}^{2} \mathrm{for} I$$

$$\alpha=\frac{0.62 \mathrm{~N} \cdot \mathrm{m}}{0.043245 \mathrm{~kg} \cdot \mathrm{m}^{2}}$$

$$=14.337 \mathrm{radian} / \mathrm{s}^{2}$$

The angular acceleration of the pulley is $$14.3 \mathrm{rad} / \mathrm{s}^{2}$$.

The angular acceleration is equal to torque divided by moment of inertia of the disc.

The angular acceleration of the pulley is $$14.3 \mathrm{rad} / \mathrm{s}^{2}$$