A torque of 12 N ∙ m is applied to a solid, uniform disk of radius 0.50 m. If the disk accelerates at 5.7 rad/s2, what is the mass of the disk?

Question

BlackTom AI · Accepted Answer

We start by identifying the rotational dynamics relation between torque, moment of inertia, and angular acceleration: τ = I α.
For a solid disk, the moment of inertia about its central axis is I = (1/2) M R^2. Here, R = 0.50 m, so I = (1/2) M (0.50)^2 = (1/2) M (0.25) = 0.125 M.
Substitute the given values into τ = I α: 12 N·m = (0.125 M)(5.7 rad/s^2).
Compute the product of the coefficient and α: 0.125 × 5.7 = 0.7125. So 12 = 0.7125 M.
Solve for M: M = 12 / 0.7125 ≈ 16.82 kg, which rounds to approximately 17 kg.
Now evaluate each option:
- Option 5 kg: If M were 5 kg, then I = 0.125 × 5 = 0.625 kg·m^2, and τ = I α = 0.625 × 5.7 ≈ 3.56 N·m, far from the given 12 N·m. This shows 5 kg cannot produce the required torque at the given acceleration.
- Option 17 kg: With M = 17 kg, I = 0.125 × 17 = 2.125 kg·m^2. Then τ = I α = 2.125 × 5.7 ≈ 12.11 N·m, which is very close to the given 12 N·m, consistent within rounding.
- Option 13 kg: I = 0.125 × 13 = 1.625 kg·m^2. τ = 1.625 × 5.7 ≈ 9.28 N·m, which is significantly less than 12 N·m, so this mass is not correct.
- Option 3 kg: I = 0.125 × 3 = 0.375 kg·m^2. τ = 0.375 × 5.7 ≈ 2.14 N·m, far too small to match the given torque and angular acceleration.
Overall, only the mass around 16.8–17 kg yields a torque consistent with the provided angular acceleration and radius, while the other masses lead to torques that are too small given τ = I α. The trend shows that increasing mass increases I and hence τ for the same α, so the incorrect options systematically underpredict the torque for the specified α.

SPHY001 Practice Quiz - Unit 10 | LA050

A torque of 12 N ∙ m is applied to a solid, uniform disk of radius 0.50 m. If the disk accelerates at 5.7 rad/s2, what is the mass of the disk?

View Explanation

Log in for full answers

Similar Questions

When a solid object rotates with a constant angular acceleration, which of the following is true?

Angular impulse is the product of Answer Question 9[input] and time. NOTE: The answer requires a single word and correct spelling. Double check the spelling.

If the four positions below were held constant and for a complete dive, which would result in the highest average angular velocity around the mediolateral axis of the body's centre of mass? Assume all other factors are the same (body mass, segment lengths and masses, dive height, etc)

Starting from rest, a 25 cm diameter disk takes 5.0 s to reach its final angular velocity of 2500 rpm (rotations per minute). Assume that the angular acceleration is constant. The disk has a moment of inertia of 2.3 × 10 − 5 kg m 2 How much torque is applied to the disk in these 5.0 s?

A solid rigid object with a moment of inertia of 12 kgm2 has an initial angular velocity of 9 rads-1 clockwise. A torque of 44 Nm was then applied to the object in the opposite direction to the initial angular velocity. After 3 s the object will have an angular speed closest to:

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler