Question

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?

Answer 1

The angular acceleration of the disk is calculated using the torque exerted by the applied force and the moment of inertia of a uniform disk. The calculated angular acceleration is 5.33 rad/s².

Step-by-step explanation:

The question involves calculating the angular acceleration of a rotating disk when a force is applied tangentially at its rim. To find the angular acceleration, we first need to calculate the torque exerted by the force and then use the moment of inertia of the disk to find the angular acceleration.

Given:

• Mass of the disk (m) = 5.0 kg
• Diameter of the disk (d) = 30 cm = 0.30 m (radius r = d/2 = 0.15 m)
• Applied force (F) = 4.0 N

The torque (τ) exerted by the force is the product of the force and the radius at which it is applied (torque = force × radius), so:

τ = F × r = 4.0 N × 0.15 m = 0.60 N·m

The moment of inertia (I) for a uniform disk is given by (1/2)mr². Thus:

I = (1/2) × 5.0 kg × (0.15 m)² = 0.1125 kg·m²

Using the formula for angular acceleration (α = torque / moment of inertia), we get:

α = τ / I = 0.60 N·m / 0.1125 kg·m² = 5.33 rad/s²

Therefore, the angular acceleration of the disk is 5.33 rad/s².

Answer 2

Angular acceleration of the disk will be
`\alpha =10.714rad/sec^2`

Step-by-step explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius
`r=(d)/(2)=(0.3)/(2)=0.15m`

Moment of inertia of disk is given by
`I=(1)/(2)mr^2=(1)/(2)* 5* 0.15^2=0.056kgm^2`

Force is given by F=4 N

Torque is given as
`\tau =Fr=4* 0.15=0.6N-m`

We also know that torque is given by
`\tau =I\alpha`

`0.6=0.056* \alpha`

`\alpha =10.714rad/sec^2`