Ch 8.2 Help
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Ch 8.2 Help
Question 1: (1 point)
A wheel starts out at rest and makes 2.763 revolutions under a constant angular acceleration of 3.076 rad/s2. It then stops abruptly. How long was the wheel in motion (see sheet 11)? Indicate with a negative (positive) sign whether the angle through which the wheel turns is (is not) given in SI units.
(How to solve):
Question 2: (1 point)
A disk under a constant angular acceleration of 0.9853 rad/s2 increases its initial angular velocity of 4.512 rads/s by a factor 2.543 . Through how many radians does the wheel turn (see sheet 7)?
(How to solve):
Question 3: (1 point)
A 0.5398 kg object (neglect its size) is held by a rod a distance of 0.5717 m from the axis of rotation and moves on a circular orbit when pushed. What is the moment of inertia of the object (see sheet 16,17)? Indicate with a positive (negative) sign whether the moment of inertia of this object increases by a factor 3 (9) when its distance to the axis of rotation is tripled.
(How to solve):
Question 4: (1 point)
A 3.988 kg uniform disk turns about an axis of rotation through its center which is perpendicular to the disk plane. The disk has a moment of inertia of 1.609 kgm2. What is the diameter of the disk (see sheet 18)? Indicate with a positive (negative) sign whether the moment of inertia with respect to an axis with the same direction but placed at the rim of the disk is smaller (larger) than for the axis in the center.
(How to solve):
A wheel starts out at rest and makes 2.763 revolutions under a constant angular acceleration of 3.076 rad/s2. It then stops abruptly. How long was the wheel in motion (see sheet 11)? Indicate with a negative (positive) sign whether the angle through which the wheel turns is (is not) given in SI units.
(How to solve):
Question 2: (1 point)
A disk under a constant angular acceleration of 0.9853 rad/s2 increases its initial angular velocity of 4.512 rads/s by a factor 2.543 . Through how many radians does the wheel turn (see sheet 7)?
(How to solve):
Question 3: (1 point)
A 0.5398 kg object (neglect its size) is held by a rod a distance of 0.5717 m from the axis of rotation and moves on a circular orbit when pushed. What is the moment of inertia of the object (see sheet 16,17)? Indicate with a positive (negative) sign whether the moment of inertia of this object increases by a factor 3 (9) when its distance to the axis of rotation is tripled.
(How to solve):
Question 4: (1 point)
A 3.988 kg uniform disk turns about an axis of rotation through its center which is perpendicular to the disk plane. The disk has a moment of inertia of 1.609 kgm2. What is the diameter of the disk (see sheet 18)? Indicate with a positive (negative) sign whether the moment of inertia with respect to an axis with the same direction but placed at the rim of the disk is smaller (larger) than for the axis in the center.
(How to solve):
guest22 Guest
Questions 1 and 2
A wheel starts out at rest and makes 2.805 revolutions under a constant angular acceleration of 2.142 rad/s2. It then stops abruptly. How long was the wheel in motion (see sheet 11)? Indicate with a negative (positive) sign whether the angle through which the wheel turns is (is not) given in SI units.
How to Solve:
Answer is POSITIVE
2.805 revolutions
angular acceleration = 2.142 (alpha)
angular motion theta = 2.805 * 2pi = 17.624 rad
theta = wt + 1/2*alpha*t^2
where w = angular velocity
t^2 = 2*theta / alpha
t = sqrt ((2*17.624)/(2.142)) = 4.0565
Question 2:
A disk under a constant angular acceleration of 1.434 rad/s2 increases its initial angular velocity of 3.028 rads/s by a factor 2.4 . Through how many radians does the wheel turn?
alpha = 1.434 rad/s^2
w = 3.028 rads/s increased by factor of 2.4
w(t) = w_o + alpha*t
w(t) = 2.4 * 3.028 = 7.2672 rads/s
t = (w_f  w_initial)/ alpha
t= (7.26723.028)/1.434 = 2.9562 sec
theta = 1/2*alpha*t^2 + w_o*t = .5(1.434)(2.9562)^2+(3.028)(2.9562) = 15.217
Anyone know how to do three and four?
How to Solve:
Answer is POSITIVE
2.805 revolutions
angular acceleration = 2.142 (alpha)
angular motion theta = 2.805 * 2pi = 17.624 rad
theta = wt + 1/2*alpha*t^2
where w = angular velocity
t^2 = 2*theta / alpha
t = sqrt ((2*17.624)/(2.142)) = 4.0565
Question 2:
A disk under a constant angular acceleration of 1.434 rad/s2 increases its initial angular velocity of 3.028 rads/s by a factor 2.4 . Through how many radians does the wheel turn?
alpha = 1.434 rad/s^2
w = 3.028 rads/s increased by factor of 2.4
w(t) = w_o + alpha*t
w(t) = 2.4 * 3.028 = 7.2672 rads/s
t = (w_f  w_initial)/ alpha
t= (7.26723.028)/1.434 = 2.9562 sec
theta = 1/2*alpha*t^2 + w_o*t = .5(1.434)(2.9562)^2+(3.028)(2.9562) = 15.217
Anyone know how to do three and four?
Kathleen Guest
Question 2
For number 2, you can just use the equation w^2=w_o^2+2a(change in theta) instead of using one to find t and then plugging that into the other equation to find theta.
guest1 Guest
Question 3
The moment of inertia is the angular mass.
The equation is simply (r^2)(m), and it's negative.
The equation is simply (r^2)(m), and it's negative.
guest1 Guest
Question 4
Here was my Question 4.
A 6.089 kg uniform disk turns about an axis of rotation through its center which is perpendicular to the disk plane. The disk has a moment of inertia of 0.711 kgm2. What is the diameter of the disk (see sheet 18)? Indicate with a positive (negative) sign whether the moment of inertia with respect to an axis with the same direction but placed at the rim of the disk is smaller (larger) than for the axis in the center.
First let's solve for true and false. Here's a little project Get a pen and hold one end either on the tip of your fingers, or lay it on a surface. The pen is the radius of a circle and where you hold it by the fingers will be the center. Hold it straight up and and with your other hand start rotating the pen, doesn't have to be far, quarter circles are fine. Now imagine on that pen two points, one at each hand, that is the center and the end. As you rotate the pen, think where the Inertia or the force is greater: towards the center or towards the end. Where does a force have to push to start the rotation?
From a mathematical stand point, the axis is also or radius. The equation for Inertia, for a Disk is I = .5*m*r^{2}, so the Inertia is proportional to the radius. If you increase the radius, what happens to the Inertia?
To Solve the actual problem;
The formula for Inertia changes slightly by a constant, we'll represent as c. The formula looks something like this: I = cMR^{2}. The constant c for a Disk (circle) is 0.5
So our formula becomes: I = .5*M*R^{2}, Now just enter the given values and use algebra.
0.711 = .5*6.089*R^{2}
0.711 = 3.0445*R^{2}
R^{2} = (0.711/3.0445)
R^{2} = 0.2335
R = 0.48326
Now it asks for the diameter, and what is the relationship between radius (r) and diameter (d)? Radius is twice the diameter (d = 2r)
2*0.48326
=
I make the final answer hidden so you can work through it and come to the answer and then check it, so spot check that and see how you do!
A 6.089 kg uniform disk turns about an axis of rotation through its center which is perpendicular to the disk plane. The disk has a moment of inertia of 0.711 kgm2. What is the diameter of the disk (see sheet 18)? Indicate with a positive (negative) sign whether the moment of inertia with respect to an axis with the same direction but placed at the rim of the disk is smaller (larger) than for the axis in the center.
First let's solve for true and false. Here's a little project Get a pen and hold one end either on the tip of your fingers, or lay it on a surface. The pen is the radius of a circle and where you hold it by the fingers will be the center. Hold it straight up and and with your other hand start rotating the pen, doesn't have to be far, quarter circles are fine. Now imagine on that pen two points, one at each hand, that is the center and the end. As you rotate the pen, think where the Inertia or the force is greater: towards the center or towards the end. Where does a force have to push to start the rotation?
From a mathematical stand point, the axis is also or radius. The equation for Inertia, for a Disk is I = .5*m*r^{2}, so the Inertia is proportional to the radius. If you increase the radius, what happens to the Inertia?
To Solve the actual problem;
The formula for Inertia changes slightly by a constant, we'll represent as c. The formula looks something like this: I = cMR^{2}. The constant c for a Disk (circle) is 0.5
So our formula becomes: I = .5*M*R^{2}, Now just enter the given values and use algebra.
0.711 = .5*6.089*R^{2}
0.711 = 3.0445*R^{2}
R^{2} = (0.711/3.0445)
R^{2} = 0.2335
R = 0.48326
Now it asks for the diameter, and what is the relationship between radius (r) and diameter (d)? Radius is twice the diameter (d = 2r)
2*0.48326
=
 Spoiler:
 0.9665
I make the final answer hidden so you can work through it and come to the answer and then check it, so spot check that and see how you do!
Guest01 Posts : 133
Join date : 20080919
#4
I did number 4 but its wrong. i get the first half of the question then im lost... any help?
new gues Guest
Re: Ch 8.2 Help
What do you mean by first half?
If it's the equation, write out equation. Show me what you are thinking.
And yes the answer for Question 4 is negative, however my purpose was to get people to think and not just copy/paste. Professor Dawber is right, even though those Aussies rarely are (joke (maybe )), there should be more reasoning. I admit it myself, I come here sometimes when I'm stuck and I can find the answer written out clear as day. I don't really learn that way, many do not as well. We've all experienced Stony Brook's science curriculum, and it's not the most person friendly. I think that if we step through this logically in a step by step process we can learn a lot more. Dawber does a great job with his hints and getting back to the students, but sometimes the hints can be more confusing, or things we know already, but stll do not know how to apply which is why we all come here.
I thought of another analogy today for the true false inertia thing. I'd be glad to explain!
If it's the equation, write out equation. Show me what you are thinking.
And yes the answer for Question 4 is negative, however my purpose was to get people to think and not just copy/paste. Professor Dawber is right, even though those Aussies rarely are (joke (maybe )), there should be more reasoning. I admit it myself, I come here sometimes when I'm stuck and I can find the answer written out clear as day. I don't really learn that way, many do not as well. We've all experienced Stony Brook's science curriculum, and it's not the most person friendly. I think that if we step through this logically in a step by step process we can learn a lot more. Dawber does a great job with his hints and getting back to the students, but sometimes the hints can be more confusing, or things we know already, but stll do not know how to apply which is why we all come here.
I thought of another analogy today for the true false inertia thing. I'd be glad to explain!
Guest01 Posts : 133
Join date : 20080919
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