Torque with Kinematic Equations: Videos & Practice Problems

In rotational dynamics, torque tau is calculated using the equation \( \tau = I \alpha \), where \( I \) is the moment of inertia and \( \alpha \) is angular acceleration. For a solid sphere, \( I = \frac{2}{5} m r^2 \). To find \( \alpha \), use the motion equation \( \omega_f = \omega_i + \alpha t \). Converting RPM to angular velocity (\( \omega \)) is essential, as is understanding the relationship between angular acceleration and torque when stopping a rotating object. This synthesis of concepts enhances problem-solving in physics.

Torque with Kinematic Equations

In rotational dynamics, understanding the relationship between torque, moment of inertia, and angular acceleration is crucial. The fundamental equation governing this relationship is given by:

Torque (τ) = I * α

Here, τ represents torque, I is the moment of inertia, and α is the angular acceleration. The moment of inertia for a solid sphere is defined as:

I = \frac{2}{5} m r^2

where m is the mass and r is the radius of the sphere. When dealing with rotational motion, it is essential to convert any given rotational speed in revolutions per minute (RPM) to angular velocity (ω) in radians per second. This conversion is done using the formula:

ω = 2πf

where f is the frequency in revolutions per second, calculated as:

f = \frac{RPM}{60}

For example, if a solid sphere has a mass of 200 kg and a diameter of 6 meters (which gives a radius of 3 meters), and it is spinning at 180 RPM clockwise, we first convert this to angular velocity:

ω = 2π \left(-\frac{180}{60}\right) = -6π \text{ radians/second}

To find the torque required to stop the sphere in a specified time, we need to determine the angular acceleration (α). The relationship between the initial and final angular velocities and angular acceleration is given by the equation:

ω_f = ω_i + αt

Rearranging this to solve for α gives:

α = \frac{ω_f - ω_i}{t}

In this scenario, if we want to stop the sphere (ω_f = 0) in 10 seconds (t = 10), we can substitute the values:

α = \frac{0 - (-6π)}{10} = \frac{6π}{10} = 1.88 \text{ radians/second}^2

Now that we have α, we can substitute it back into the torque equation. First, we calculate the moment of inertia:

I = \frac{2}{5} (200) (3^2) = \frac{2}{5} (200) (9) = 720 \text{ kg m}^2

Finally, substituting I and α into the torque equation gives:

τ = I * α = 720 * 1.88 = 1353.6 \text{ Newton meter}

This example illustrates the interconnectedness of torque, angular acceleration, and moment of inertia in solving rotational motion problems. By systematically applying the relevant equations and converting units as necessary, one can effectively analyze and solve complex rotational dynamics scenarios.

A light, long rope is wrapped around a solid disc, in such a way that pulling the rope causes the disc to spin about a fixed axis perpendicular to itself and through its center. The disc has mass 40 kg, radius 2 m, and is initially at rest, and the rope unwinds without slipping. You pull on the rope with a constant 200 N. Use the rotational version of Newton's Second Law to calculate how fast (in rad/s) the disc be spinning after you pull 50 m of rope.

• A. 3.2 rad/s
• B. 15.8 rad/s
• C. 22.4 rad/s
• D. 633 rad/s

Answer: (B)

A system is made of two small, 3 kg masses attached to the ends of a 5 kg, 4 m long, thin rod, as shown. The system is free to rotate about an axis perpendicular to the rod and through its center. Two forces, both of magnitude F and perpendicular to the rod, are applied as shown below. What must the value of F be to the system from rest to 10 rad/s in exactly 8 complete revolutions?

• A. 3.8 N
• B. 6.0 N
• C. 7.7 N
• D. 15.4 N

Answer: (C)

Two rotating doors, each 6.0 m long, are fixed to the same central axis of rotation, as shown (top view). When you push on one door with a constant 100 N, directed perpendicular from the face of the door and 50 cm from its outer edge, the rotating door system takes 8 s to complete a full revolution from rest. The doors can be modeled as thin rectangles (moments of inertia for thin rectangles, around two different axes, are shown for reference). Calculate the mass of the system.

• A. 21 kg
• B. 84 kg
• C. 104 kg
• D. 416 kg

Answer: (D)

Stopping flywheel with friction

In this scenario, we analyze a flywheel, which is a rotating disc, and determine the force required to stop it within a specified time frame. The moment of inertia for a hollow disc is given by the formula \( I = m r^2 \), where \( m \) is the mass and \( r \) is the radius. For our flywheel, with a mass of \( 8 \times 10^4 \) kg and a diameter of 5 meters (thus a radius of 2.5 meters), we can calculate its moment of inertia.

To bring the flywheel to a complete stop, we need to apply a force against it. The initial angular velocity (\( \omega \)) of the flywheel is 300 RPM, which we convert to radians per second using the formula \( \omega = \frac{2\pi \text{RPM}}{60} \), resulting in \( \omega = 10\pi \) radians per second. The stopping time (\( \Delta t \)) is 30 seconds, and we need to find the angular acceleration (\( \alpha \)) required to achieve this.

Using the equation for angular motion, \( \omega_f = \omega_i + \alpha t \), where \( \omega_f \) is the final angular velocity (0 in this case), we can rearrange to find \( \alpha \). Plugging in the values gives us \( \alpha = -\frac{\omega_i}{t} = -\frac{10\pi}{30} = -\frac{\pi}{3} \) radians per second squared, which simplifies to approximately \( -1.05 \) radians per second squared.

The torque (\( \tau \)) generated by the friction force is given by \( \tau = I \alpha \). The friction force can be expressed as \( f_{\text{friction}} = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. In this case, the static coefficient of friction is 0.8, and the kinetic coefficient is 0.6. Since the block is sliding against the disc, we use the kinetic coefficient of friction.

Substituting the expressions for torque and friction into the equation \( \tau = f_{\text{friction}} \cdot r \), we can derive the force needed to stop the flywheel. The equation becomes \( f_{\text{friction}} \cdot r = I \alpha \). By substituting \( f_{\text{friction}} = \mu f \) and \( I = m r^2 \), we can solve for the force \( f \). This leads to the equation:

\( f = \frac{I \alpha}{\mu} \)

After substituting the known values, we find that the required force to stop the flywheel in 30 seconds is approximately \( 3.5 \times 10^5 \) newtons. This calculation illustrates the relationship between force, torque, and angular acceleration in rotational dynamics, emphasizing the importance of understanding how these concepts interconnect to solve real-world problems.

A 1,000 kg disc that has a 5 m outer radius is mounted on a vertical, inner axle 80 kg in mass and 1 m in radius. A motor acts on the axle to speed up or slow down the system. Suppose the motor stops functioning when the system is spinning at 70 rad/s. To bring it to a complete stop, you apply a constant 200 N friction to the surface of the axle. How many revolutions will the system take to stop?

• A. 5.8
• B. 2.6×10⁴
• C. 1.6×10⁵
• D. 2.5×10⁵

Answer: (B)

Here’s what students ask on this topic:

How do you calculate torque in rotational dynamics?

In rotational dynamics, torque (τ) is calculated using the equation $\tau =I\alpha$, where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. The moment of inertia depends on the shape and mass distribution of the object. For example, for a solid sphere, $I=\frac{2}{5}m{r}^2$. Angular acceleration can be found using kinematic equations such as $\omega f_{}={\omega }_i+\alpha t$.

What is the relationship between torque and angular acceleration?

The relationship between torque (τ) and angular acceleration (α) is given by the equation $\tau =I\alpha$, where $I$ is the moment of inertia. This equation is the rotational analog of Newton's second law of motion (F = ma). It shows that the torque applied to an object is directly proportional to its angular acceleration, with the moment of inertia acting as the proportionality constant.

How do you convert RPM to angular velocity?

To convert RPM (revolutions per minute) to angular velocity (ω) in radians per second, use the formula $\omega =2\pi \left(\frac{\mathrm{RPM}}{60}\right)$. This conversion is necessary because most rotational motion equations use angular velocity in radians per second rather than RPM.

How do you find angular acceleration using kinematic equations?

Angular acceleration (α) can be found using the kinematic equation $\omega f_{}={\omega }_i+\alpha t$. Rearrange this equation to solve for α: $\alpha =\frac{{\omega }_f-{\omega }_i}{t}$. Here, ${\omega }_f$ is the final angular velocity, ${\omega }_i$ is the initial angular velocity, and $t$ is the time interval.

What is the moment of inertia for a solid sphere?

The moment of inertia (I) for a solid sphere rotating about an axis through its center is given by the formula $I=\frac{2}{5}m{r}^2$, where $m$ is the mass of the sphere and $r$ is its radius. This formula is specific to solid spheres and differs for other shapes and mass distributions.