[table]
In Professor Oak’s lab, a new Poké Ball launcher is being tested. The launcher uses a spinning platform shaped like a solid disk, with a mass of 2.90 kg and a diameter of 0.300 m. When powered on, a motor applies a constant torque, bringing the platform from rest to an angular speed of 1600 rpm in 4.2 seconds. How much work does the motor do during this spin-up process?
[/table]

Question

BlackTom AI · Accepted Answer

Question restatement: A solid-disk platform with mass m = 2.90 kg and diameter D = 0.300 m (radius r = 0.150 m) is spun up from rest to an angular speed ω = 1600 rev/min in t = 4.2 s by a motor applying a constant torque. We are asked for the work done by the motor during this spin-up.

Key physics steps to analyze:
- For a solid disk, the moment of inertia is I = (1/2) m r^2.
- Convert the final angular speed to radians per second: 1600 rev/min × (2π rad/rev) × (1 min/60 s).
- The angular acceleration α = Δω / Δt, though for work we can directly use the rotational kinetic energy: W = ΔK = (1/2) I ω^2 (since starting from rest, ΔK = K_final).
- Compute ω in rad/s, then evaluate I, then plug into W = (1/2) I ω^2.

Option-by-option analysis:
- Option a: 339 J. If this were correct, it would imply a smaller kinetic energy at the end of spin-up. Given the disk’s inertia (I) and the fairly large final angular velocity, a 339 J value would require either a smaller I or a smaller ω than the actual values. Since I depends on m and r^2, and ω is fixed by the problem, this option underestimates the energy stored in rotation and is not consistent with the full (1/2) I ω^2 calculation for the stated final speed.
- Option b: 414 J. This is closer to the expected value but still lower than the precise calculation. The discrepancy could come from rounding in intermediate steps, but using exact I and ω yields a somewhat larger result than this, so this option is not the exact value from the standard rotational-work computation.
- Option c: 458 J. This option is very near the value obtained from a careful calculation. To check, compute I = (1/2) m r^2 = 0.5 × 2.90 × (0.150)^2 ≈ 0.032625 kg·m^2. Convert ω = 1600 rev/min to rad/s: ω = 1600 × 2π / 60 ≈ 167.55 rad/s. Then W = (1/2) I ω^2 ≈ 0.5 × 0.032625 × (167.55)^2 ≈ 0.0163125 × 28100 ≈ 458 J. This aligns well with option 458 J, making it a strong candidate for the correct numerical result.
- Option d: 399 J. This value is somewhat below the precise rotational energy. It would imply a smaller ω or I than given, or a miscalculation in squaring ω or in the inertia term. The standard expression yields a number noticeably different from 399 J when computed with the provided m and r.
- Option e: 327 J. This is significantly lower than the expected rotational energy for the stated final speed and given disk parameters. Like the others, matching this would require deviating from the problem’s stated m, r, and ω, so this option is not consistent with the correct kinetic-energy evaluation.

Additional considerations and checks:
- Using the exact I for a solid disk and the exact ω reduces rounding errors; the computed W ≈ 458 J matches option c when kept precise, whereas the other options correspond to underestimates that would result from rounding or misapplication of the formula.
- The torque being constant is consistent with a constant angular acceleration, but for work we only need the final rotational energy since the starting kinetic energy is zero; the work-energy theorem in rotational form gives W = ΔK = (1/2) I ω^2 regardless of the path, as long as ω starts at 0 and ends at the given value.
- There is no need to invoke linear work or translational energy here because the system is purely rotational about the disk's center of mass, and the axis is fixed for the spinning disk.

Summary assessment of each option:
- a. 339 J: too small; underestimates W compared to the full rotational energy by the computed ω and I.
- b. 414 J: closer but still below the precise calculation; not exact for the given numbers.
- c. 458 J: matches the exact rotational-energy calculation using I and ω from the problem, making it the most accurate option.
- d. 399 J: underestimates, similar in direction to a and b but farther from the precise value.
- e. 327 J: significantly lower than the computed value; not consistent with the correct energy budget.

View Explanation

Log in for full answers

Similar Questions

What is a rotational kinetic energy and what is its SI unit?

An ice skater performs a fast spin by pulling outstretched arms down and close to her body. What happens to her kinetic energy with respect to the axis of rotation?

Suppose a solid uniform cylinder rolls without slipping down an inclined plane (like in Figure 12.42). Once the object starts rolling then the rotational kinetic energy of the cylinder

A system consists of a 2.0 kg uniform disk with radius 0.40 m, and ring of negligible thickness and mass 1.0 kg with a radius of 0.20 m, as shown. If the system has a rotational speed of 50.0 rad/s, what is its kinetic energy in Joules?

In a consumer society, many adults channel creativity into buying things

Economic stress and unpredictable times have resulted in a booming industry for self-help products

People born without creativity never can develop it

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