Question

A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight $\mathrm{mg}$ is attached to the end of the cord and falls from the rest. After falling through the distance $\mathrm{h}$. the angular velocity of the wheel will be.... $$\begin{gathered} B \\ \left[2mgh/(I+m{r}^2)\right] \end{gathered}$$ $$\begin{gathered} \mathrm{C} \\ {\left[2\mathrm{mgh}/(\mathrm{I}+{\mathrm{mr}}^2)\right]}^{1/2} \end{gathered}$$ $$\begin{gathered} \mathrm{D} \\ \sqrt{(2\mathrm{gh})} \end{gathered}$$

Short answer

The angular velocity of the wheel after the weight falls from rest is given by $\omega =\sqrt{\frac{2mgh}{m{r}^2+I}}$, which corresponds to option $\boxed{{\displaystyle C{\left[2\mathrm{mgh}/(\mathrm{I}+{\mathrm{mr}}^2)\right]}^{1/2}}}$.

Step-by-step solution

Calculate the initial mechanical energy

First, we need to find the initial mechanical energy, which is the potential energy of the weight. The potential energy can be calculated as: $$U_{initial}=mgh$$

Calculate the final mechanical energy

When the weight falls, the wheel gains kinetic energy from the weight's potential energy. The kinetic energy will be split into two parts: linear kinetic energy of the weight and rotational kinetic energy of the wheel. At the final point, the potential energy will be zero. Linear kinetic energy of the weight: $$K_{linear}=\frac{1}{2}m{v}^2$$ Rotational kinetic energy of the wheel: $$K_{rotational}=\frac{1}{2}I{\omega }^2$$ Total kinetic energy (as potential energy has been converted to kinetic energy): $$U_{final}=K_{linear}+K_{rotational}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$

Conserve mechanical energy

The sum of all initial energies should equal the sum of all final energies. Thus, we have the following equation: $$U_{initial}=U_{final}$$ $$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$

Relate the linear velocity of the weight to the angular velocity

The linear velocity of the weight can be related to the angular velocity of the wheel by the following equation: $$v=r\omega$$

Substitute linear velocity in the conservation of energy equation

We substitute the linear velocity with the angular velocity equation from step 4 to get: $$mgh=\frac{1}{2}m(r\omega {)}^2+\frac{1}{2}I{\omega }^2$$

Simplify and solve for the angular velocity

Now, we need to simplify and solve for the angular velocity of the wheel: $$mgh=\frac{1}{2}m{r}^2{\omega }^2+\frac{1}{2}I{\omega }^2$$ $$2mgh=(m{r}^2+I){\omega }^2$$ Finally, solving for $\omega$, we get: $$\omega =\sqrt{\frac{2mgh}{m{r}^2+I}}$$ This matches option C: $$\boxed{{\displaystyle C{\left[2\mathrm{mgh}/(\mathrm{I}+{\mathrm{mr}}^2)\right]}^{1/2}}}$$

Key concepts

Mechanical Energy Conservation

When an object is in motion, its energy is constantly exchanged between different forms. One of the key principles of physics is the conservation of mechanical energy. Mechanical energy can appear as potential energy or kinetic energy. In a closed system where no energy is lost to friction or air resistance, the total mechanical energy remains constant.
Let's break this down:

• Potential energy (U_{initial}) is the energy stored in an object due to its position. For a weight that is elevated at height h, the potential energy is calculated as $U_{initial}=mgh$.
• As the weight falls, this stored potential energy is transformed into kinetic energy, specifically linear kinetic energy of the falling weight ($K_{linear}=\frac{1}{2}m{v}^2$) and rotational kinetic energy of the wheel ($K_{rotational}=\frac{1}{2}I{\omega }^2$).

By the conservation of mechanical energy principle, the total energy initially in the form of potential energy gets converted into these kinetic energy forms. The equation becomes: $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$. This relationship allows us to equate different forms of energy to solve for other unknowns, like angular velocity.

Moment of Inertia

In rotational motion, moment of inertia plays a role similar to mass in linear motion. It is a measure of how difficult it is to change the rotational motion of an object. The moment of inertia, denoted by I, depends on the mass of the object and the distribution of mass around the axis of rotation.
Think of it this way:

• A compact disc that spins might rotate easily because its mass is concentrated near its center, leading to a smaller moment of inertia.
• A solid cylinder with the same mass but larger radius has more mass distributed further away, giving it a larger moment of inertia.

In our exercise, the moment of inertia of the wheel affects the rate at which it can spin when the weight moves its potential energy to kinetic energy. The greater I is, the slower it accelerates for a given torque.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. Similar to linear kinetic energy which depends on mass and velocity, rotational kinetic energy is dependent on moment of inertia and angular velocity.
The formula for rotational kinetic energy is given by:$K_{rotational}=\frac{1}{2}I{\omega }^2$, where $\omega$ represents the angular velocity.
As the weight falls, it causes the wheel to spin, transferring potential energy into rotational kinetic energy. The amount of energy transferred into spinning the wheel can be calculated using the moment of inertia. It highlights how an object's frame of rotation influences its energy and speed.

Linear Velocity

The linear velocity of an object is the rate of change of its position with respect to time and is directed along a straight path. For a rotating object, like a wheel, linear and angular velocity are closely related through the radius of the rotation.
The relationship is described by the equation: $v=r\omega$, where v is the linear velocity, r is the radius, and $\omega$ is the angular velocity.
In the scenario presented, when the weight falls and causes the wheel to spin, the linear velocity of the weight is directly proportional to the angular velocity of the wheel. This link helps us to compute the speed of fall and rotation from one another, highlighting how energy is effectively transferred between different types of motion.