A thin rod of uniform density with mass
𝑀
=
1.0
kg
and length
𝐿
=
0.80
m
can rotate freely around the pivot O, as shown in the diagram. The other end of the rod is held by a hand such that the rod is horizontal. When the right end of the rod is released, what is the angular acceleration (in units of
rad/s
2
) of the rod at this instant?
[ Select ]
36
27
6.1
18
72

When the rod swings to the vertical position, as shown in the diagram, what is the angular speed (in units of rad/s) of the rod at this instant?
[ Select ]
3.3
14
6.1
1.8
2.4

Question

A thin rod of uniform density with mass 
𝑀
=
1.0
 kg
 and length 
𝐿
=
0.80
 m
 can rotate freely around the pivot O, as shown in the diagram. The other end of the rod is held by a hand such that the rod is horizontal. When the right end of the rod is released, what is the angular acceleration (in units of 
rad/s
2
) of the rod at this instant? 
[ Select ]
36
27
6.1
18
72

When the rod swings to the vertical position, as shown in the diagram, what is the angular speed (in units of rad/s) of the rod at this instant? 
[ Select ]
3.3
14
6.1
1.8
2.4

BlackTom AI · Accepted Answer

To tackle the two parts, I’ll break down the physics clearly and then evaluate each option step by step.

First, establish the system data: a uniform rod of mass M = 1.0 kg and length L = 0.80 m is hinged at O at one end. The rod starts horizontal with the right end released.

Option set for the first question (angular acceleration when released):
- 36 rad/s^2
- 27 rad/s^2
- 6.1 rad/s^2
- 18 rad/s^2
- 72 rad/s^2

Analysis of each option for the initial angular acceleration:
- 36: If you compute alpha = τ/I using torque about the pivot, you need τ = Mg(L/2) = (1.0 kg)(9.8 m/s^2)(0.40 m) = 3.92 N·m and I = (1/3)ML^2 = (1/3)(1.0 kg)(0.80 m)^2 = 0.213... kg·m^2. Then alpha = 3.92 / 0.213... ≈ 18.4 rad/s^2. This is far from 36, so 36 is too large.
- 27: This would require a larger torque or a smaller moment of inertia than our rigid-rod calculation provides. Since the geometry and mass are fixed, 27 is not consistent with τ/I = Mg(L/2) / (1/3 ML^2). So it’s incorrect.
- 6.1: This is about one third of the correct value; the torque-to-inertia ratio for this setup does not yield such a small alpha, so 6.1 is not correct for the initial release.
- 18: This is the closest to the correct result when using α = τ/I with τ = Mg(L/2) and I = (1/3)ML^2. Substituting, α ≈ 3.92 / 0.213... ≈ 18.4 rad/s^2, i.e., about 18 rad/s^2. Given the approximate g and rounding, this matches.
- 72: This is clearly too large for the described system; it would imply an enormous torque-to-inertia ratio that isn’t supported by the values of M, g, L, and the rod’s I. So it’s incorrect.

Next, analyze the second question: angular speed when the rod reaches the vertical position.
- 3.3 rad/s
- 14 rad/s
- 6.1 rad/s
- 1.8 rad/s
- 2.4 rad/s

We use energy conservation from the initial horizontal position (the right end released) to the vertical bottom:
- The change in gravitational potential energy as the center of mass drops by h = L/2 = 0.40 m is ΔU = Mg h = (1.0 kg)(9.8 m/s^2)(0.40 m) = 3.92 J.
- This converts into rotational kinetic energy: (1/2) I ω^2, with I = (1/3) ML^2 = 0.213... kg·m^2.
- Set Mg h = (1/2) I ω^2 and solve for ω: ω^2 = 2 M g h / I = 2(3.92) / 0.213... ≈ 7.84 / 0.213... ≈ 36.7, so ω ≈ 6.1 rad/s.

Thus, the second answer matches 6.1 rad/s.

Summary of reasoning progression:
- For the first part, compute I about the hinge and the gravitational torque about the hinge, then α = τ/I. The numeric evaluation aligns with ~18 rad/s^2, making 18 the correct choice among the options.
- For the second part, apply energy conservation from horizontal to vertical, equate potential energy loss to rotational kinetic energy, and solve for ω. The result is about 6.1 rad/s, matching that option.

View Explanation

Log in for full answers

Similar Questions

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?

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