A disk with a rotational inertia of 2.0 kg.m^2 and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 5.0 N is applied tangentially to the rim. The angular acceleration of the disk is:

Question

A disk with a rotational inertia of 2.0 kg.m^2 and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 5.0 N is applied tangentially to the rim. The angular acceleration of the disk is:

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4 answers
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Answer 1

jay answered
10 years ago

never mind i got it.

Answer 2

jh answered
4 years ago

what is the answer pls reply asap

Answer 3

Step-by-Step Bot answered
8 months ago

To find the angular acceleration of the disk, we can use Newton's second law for rotational motion, which states that the torque applied to an object is equal to the product of the moment of inertia and the angular acceleration.

The torque (τ) acting on the disk can be calculated using the formula: τ = r x F, where r is the radius of the disk and F is the force applied tangentially to the rim.

Given that the radius (r) of the disk is 0.40 m and the force (F) applied is 5.0 N, we can calculate the torque:

τ = r x F
= 0.40 m x 5.0 N
= 2.0 N.m

Now, we can use the formula τ = I α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Plugging in the values:

2.0 N.m = 2.0 kg.m^2 x α

Simplifying the equation:

α = 2.0 N.m / 2.0 kg.m^2

α = 1.0 rad/s^2

Therefore, the angular acceleration of the disk is 1.0 rad/s^2.

Answer 4

Explain Bot answered
8 months ago

To find the angular acceleration of the disk, we can use the formula:

τ = Iα

Where:
τ is the torque applied to the disk
I is the rotational inertia of the disk
α is the angular acceleration of the disk

First, we need to find the torque applied to the disk. The torque is given by the formula:

τ = rFsin(θ)

Where:
r is the radius of the disk
F is the applied force
θ is the angle between the force and the radius vector

In this case, the applied force is tangentially applied to the rim, so the angle θ between the force and the radius will be 90 degrees.

τ = rFsin(90°)
= rF

Substituting the known values:
τ = (0.40 m)(5.0 N)
= 2.0 N·m

Now we can use the torque value and the rotational inertia given to find the angular acceleration.

2.0 N·m = (2.0 kg·m^2)·α

Rearranging the equation to solve for α:

α = 2.0 N·m / (2.0 kg·m^2)
= 1.0 N·m / kg·m^2

Therefore, the angular acceleration of the disk is 1.0 N·m/kg·m^2.