Need help with torque and angular inertial problems

AI Thread Summary
The discussion focuses on two physics problems involving torque and angular inertia. In the first problem, two particles are attached to a massless rod, and participants are confused about how to calculate the initial accelerations of the particles using angular inertia, questioning whether mass should be neglected. The second problem involves two blocks connected by a cord around a disk, where the correct acceleration of the blocks is calculated as 2.635 m/s², but participants seek clarification on the steps taken to arrive at this answer. Key points include the application of Newton's second law for rotational motion and the relationship between linear and angular acceleration. Overall, the conversation emphasizes the importance of understanding mass and inertia in solving these types of physics problems.
Shadow Cloud
Messages
13
Reaction score
0
Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released.
hrw7_10-43.gif

a.What is the magnitude of the initial acceleration of particle 1?
b.What is the magnitude of the initial acceleration of particle 2?

I have no idea where to begin with this problem when the only thing they give me is length. Should mass be neglected in the equation for angular inertia?
In Figure 10-53, two blocks, of mass m1 = 310 g and m2 = 630 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.
hrw7_10-53.gif

Find the magnitude of the acceleration of the blocks.
a= 2.635 m/s^2 (I got the right answer, but I don't specifically understand the steps to getting it)
b. Find the tension of T1
c. Find the tensin of T2

For the last problem in finding the tension, why does the following not work?
-Fg + T1 = ma
T1 = Fg + ma
T1 = (.310 kg * 9.8m/s^2) + (.310 kg * 2.635 m/s^2)
 
Physics news on Phys.org
Shadow Cloud said:
Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released.
hrw7_10-43.gif

a.What is the magnitude of the initial acceleration of particle 1?
b.What is the magnitude of the initial acceleration of particle 2?

I have no idea where to begin with this problem when the only thing they give me is length. Should mass be neglected in the equation for angular inertia?
How can you ignore the mass when talking about inertia?!
Apply Newton's 2nd law for rotational motion: \tau = I \alpha
In Figure 10-53, two blocks, of mass m1 = 310 g and m2 = 630 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.
hrw7_10-53.gif

Find the magnitude of the acceleration of the blocks.
a= 2.635 m/s^2 (I got the right answer, but I don't specifically understand the steps to getting it)

Show how you solved for the acceleration.
b. Find the tension of T1
c. Find the tensin of T2
For the last problem in finding the tension, why does the following not work?
-Fg + T1 = ma
T1 = Fg + ma
T1 = (.310 kg * 9.8m/s^2) + (.310 kg * 2.635 m/s^2)
Assuming you have the correct acceleration (I haven't checked) then that should give you the correct tension. Show how you solved for the acceleration.
 
Last edited:
Thank you for the reply. You are right though, I did get the correct answer for the tension. I have no idea what I did wrong before. I must had punched in the wrong values or something.
How did I get the acceleration? Professly, I would like to understand this better. I originally got help with this problem, but the person never really explained his steps.
F = ma
F = (m2 - m1)
F = (M1 + M2)a + Ia (angular) / R I'm unsure how Ia/R comes into play
(m1 + m2 + 0.5M)a = (m2 - m1)g
a = (g(m2 -m1)) / (m1 + m2 + .5M)
a = 2.635 m/s^2


I know you need mass for t = Ia , but the fact that there is no value for m is what makes me confused as to what to do. I am surmizing that 9.8 m/s^2 factors into that problem somehow?
 
That's one way to solve it, but I don't recommend it. Instead I'd ask you to analyze each object separately, applying Newton's 2nd law to each. You'll end up with 3 equations (one for each mass and one for the disk) that you can solve for a, T1, & T2. Give it a shot and see how you do.

You'll need to use a = \alpha R, to relate the linear and angular accelerations. (9.8 m/s^2 is the acceleration due to gravity!) You are given the masses of each object. (What's the rotational inertia of a disk?)
 
Oh I'm sorry by my last statement of "I know you need mass for t = Ia , but the fact that there is no value for m is what makes me confused as to what to do. I am surmizing that 9.8 m/s^2 factors into that problem somehow?"
I was referring to this problem.
Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released.
hrw7_10-43.gif

a.What is the magnitude of the initial acceleration of particle 1?
b.What is the magnitude of the initial acceleration of particle 2?
 
You are given all the mass there is. (The rod has no mass; all the mass is at the ends of the rod.) What's the rotational inertia of a mass at the end of massless rod? (Hint: All the mass is concentrated at a single point. How far is the mass from the axis of rotation?)

And 9.8 m/s^2 is still the acceleration due to gravity! Of course you'll need it.
 
I still don't have a clue as to what I should do. Okay providing I'm on the right track. I should start out with...
T=Ia
T = mr^2a
a = T / mr^2


The problem is I don't know what else to do from here. Since the rod is massless does this mean that m is ignored or something? There's no force mentioned in the problem so that the net torque can be solved. The only thing I have is the radius.
 
Shadow Cloud said:
I should start out with...
T=Ia
T = mr^2a
a = T / mr^2
You are on the right track. I'll rewrite that as follows so as not to confuse angular acceleration with linear acceleration:
\tau = I \alpha

One mass is a distance L_1 from the axis; the other, L_2. Find the total rotational inertia of both masses.
The problem is I don't know what else to do from here. Since the rod is massless does this mean that m is ignored or something?
The mass of the rod is assumed to be negligible; but it does serve to hold the two masses (m) in place.
There's no force mentioned in the problem so that the net torque can be solved.
If there's mass, then there's gravity.

Once you find the angular acceleration, you can find the linear acceleration using:
a = \alpha R
 

Similar threads

Replies
30
Views
3K
Replies
4
Views
1K
Replies
10
Views
2K
Replies
22
Views
4K
Replies
5
Views
5K
Replies
1
Views
1K
Replies
4
Views
13K
Replies
4
Views
1K
Back
Top