Lab Quiz 6
Page 1 of 2
Page 1 of 2 • 1, 2
Lab Quiz 6
Question 1: (1 point)
You spin a wheel to give it an initial angular acceleration. The angular velocity of the wheel decreases with time due to a frictional torque acting on the axle of the wheel. You plot the angular velocity as a function of time. When determining the slope of the linear graph, you measure a change of the angular velocity with a magnitude of 3.5 rad/s for a time interval of 11.4 s. What is the magnitude of the angular acceleration due to friction (see Ch8 sheet 5)? Indicate with positive (negative) sign whether the sign of the measured acceleration is negative (positive).
(How to solve):
Question 2: (1 point)
You calculate the moment of inertia of a disk with a handle attached as shown above. The mass of the disk + handle is 4.9 kg and the radius of the disk is 35.4 cm. You use the equation in Ch8 sheet 18, which holds for the moment of inertia of a uniform disk, and ignore the handle. What is the moment of inertia? Indicate with a negative (positive) sign whether you overestimate (underestimate) the moment of inertia of disk + handle, and the error you make is small (large). (Hint: when you judge whether the error made is small or large, consider the mass and the radius of the handle relative to the disk).
(How to solve):
Question 3: (1 point)
You measure the moment of inertia of a wheel by measuring its angular acceleration a = 0.413 rad/s2, when applying an external torque caused by a hanging weight of mass m = 200 g as shown above. The angular acceleration is decreased by a frictional torque, accounted for by afric = - 0.193 rad/s2 in the expression for the momemt of inertia I of the wheel, I = mr(g-ra) /(|afric|+a), where r is the radius of the cylinder attached to the wheel, and g is the acceleration of gravity, which you treat as error free. You neglect the error of the accelerations a and afric, and the error of the mass m. You find out that the term (a r) in the numerator is small compared to the term g, and thus neglect its error too. What is the absolute error of I, if the 2.5 cm radius r is known with a 1.3 % accuracy. (Hint: Read in the manual of the lab, how I can be written as two factors. One factor contains the quantities assumed to be error free, after taking into account that the term (a r) can be neglected for the purpose of error calculation. Use expression(3) and (4) in "Error and Uncertainty".) Indicate with a positive (negative) sign whether, with the neglections you made above, the absolute (relative) errors of I and r are the same.
(How to solve):
You spin a wheel to give it an initial angular acceleration. The angular velocity of the wheel decreases with time due to a frictional torque acting on the axle of the wheel. You plot the angular velocity as a function of time. When determining the slope of the linear graph, you measure a change of the angular velocity with a magnitude of 3.5 rad/s for a time interval of 11.4 s. What is the magnitude of the angular acceleration due to friction (see Ch8 sheet 5)? Indicate with positive (negative) sign whether the sign of the measured acceleration is negative (positive).
(How to solve):
Question 2: (1 point)
You calculate the moment of inertia of a disk with a handle attached as shown above. The mass of the disk + handle is 4.9 kg and the radius of the disk is 35.4 cm. You use the equation in Ch8 sheet 18, which holds for the moment of inertia of a uniform disk, and ignore the handle. What is the moment of inertia? Indicate with a negative (positive) sign whether you overestimate (underestimate) the moment of inertia of disk + handle, and the error you make is small (large). (Hint: when you judge whether the error made is small or large, consider the mass and the radius of the handle relative to the disk).
(How to solve):
Question 3: (1 point)
You measure the moment of inertia of a wheel by measuring its angular acceleration a = 0.413 rad/s2, when applying an external torque caused by a hanging weight of mass m = 200 g as shown above. The angular acceleration is decreased by a frictional torque, accounted for by afric = - 0.193 rad/s2 in the expression for the momemt of inertia I of the wheel, I = mr(g-ra) /(|afric|+a), where r is the radius of the cylinder attached to the wheel, and g is the acceleration of gravity, which you treat as error free. You neglect the error of the accelerations a and afric, and the error of the mass m. You find out that the term (a r) in the numerator is small compared to the term g, and thus neglect its error too. What is the absolute error of I, if the 2.5 cm radius r is known with a 1.3 % accuracy. (Hint: Read in the manual of the lab, how I can be written as two factors. One factor contains the quantities assumed to be error free, after taking into account that the term (a r) can be neglected for the purpose of error calculation. Use expression(3) and (4) in "Error and Uncertainty".) Indicate with a positive (negative) sign whether, with the neglections you made above, the absolute (relative) errors of I and r are the same.
(How to solve):
hwilson- Guest
Re: Lab Quiz 6
Question 1:
Answer is positive
angular velocity / time
Question 2:
Answer is negative
.5 (constant for a disc) * mass * radius(in meters)^2
Any ideas for Question 3?
Answer is positive
angular velocity / time
Question 2:
Answer is negative
.5 (constant for a disc) * mass * radius(in meters)^2
Any ideas for Question 3?
Andrew- Guest
question 3
answer: negative
after you done all error neglect and calculate out the I
the error equation should be
DeltaI(absolute error of I)=sqrt((Deltar/r)^2)*I
DeltaI=Deltar/r*I I is calulated through m*r*(g-ar)/(afric+a), where afric is absolute vaule, ar=angular accelerration * radius
the answer is negative because DeltaI*I=Deltar*r
after you done all error neglect and calculate out the I
the error equation should be
DeltaI(absolute error of I)=sqrt((Deltar/r)^2)*I
DeltaI=Deltar/r*I I is calulated through m*r*(g-ar)/(afric+a), where afric is absolute vaule, ar=angular accelerration * radius
the answer is negative because DeltaI*I=Deltar*r
chen- Guest
question 3
The answer works on the CD practice question, but it is not working on maple TA, i don't know why, i try many times already, still not corrected.
does anyone have any idea?
does anyone have any idea?
chen- Guest
question 3
I keep getting question three wrong could someone please post a step by step procedure for it?
hwilson- Guest
has anyone gotten #3 correct?
Ive tried question 3 using
deltar/r = deltaI/I
as well as just deltar/r and neither came out right... (i put answers in as negative)
does anyone know how to do this problem/has anyone gotten it correct on mapleTA? thanks!
deltar/r = deltaI/I
as well as just deltar/r and neither came out right... (i put answers in as negative)
does anyone know how to do this problem/has anyone gotten it correct on mapleTA? thanks!
phyguest- Guest
Re: Lab Quiz 6
I am still having trouble with Question 3 as well.
Guest01- Posts : 133
Join date : 2008-09-19
Re: Lab Quiz 6
can someone who had #3 correct post exactly what they did because the one posted above is not understandable..thanks
Question- Guest
Re: Lab Quiz 6
it says to refer to the lab procedure but it is not ready yet (on black board).
waiting- Guest
question3
question3
multiply the error of r they give(over 100) by I, which you work out using the formula given
multiply the error of r they give(over 100) by I, which you work out using the formula given
stranger- Guest
ques 3
Number 3 still wont work for me, can you help me figure out what i did wrong?
angular acc = .388
afric = -.171
r=.025 m
m= .2kg
error of r = 1.35%
1.35/100 = .0135
I = mr (g-r*angularacc) / (abs afric + angular acc)
I= (.2)(.025) * (9.81-(.025*.388)) / (.171 + .388)
= .087659
.087659 * .0135 = .0011833
and i put it as negative. any suggestions?
angular acc = .388
afric = -.171
r=.025 m
m= .2kg
error of r = 1.35%
1.35/100 = .0135
I = mr (g-r*angularacc) / (abs afric + angular acc)
I= (.2)(.025) * (9.81-(.025*.388)) / (.171 + .388)
= .087659
.087659 * .0135 = .0011833
and i put it as negative. any suggestions?
confused- Guest
ques 3
ive tried it with both grams and kg so idk which is right, and are we sure its negative?
k- Guest
Re: Lab Quiz 6
Strangers worked for me. Symbols they use for the variables in the question is confusing tho...
guesto- Guest
Re: Lab Quiz 6
and in reply to confused, maybe you had rounded off too early because the answer I got at the end without rounding off before hand is .001182
guesto- Guest
answer to question 3
The following steps worked out for me. This is what I did:
Wanted: Delta I= I * Delta r/ r
First find I=mg/(alpha fric+alpha) Note: take the absolute value of alpha friction.
Then, find Deta r/r = (% accuracy/100 * radius
Finally, multiply two values together.
(it's negative)
Hope it helps and will work for you guys!
Wanted: Delta I= I * Delta r/ r
First find I=mg/(alpha fric+alpha) Note: take the absolute value of alpha friction.
Then, find Deta r/r = (% accuracy/100 * radius
Finally, multiply two values together.
(it's negative)
Hope it helps and will work for you guys!
tofu- Guest
ques 3
Yeah, tofu's suggestion for question 3 works, and remember to put the mass in kg!
Kathleen- Guest
Re: Lab Quiz 6
I am still having trouble with Question 3
can someone plug in the numbers so we get it
can someone plug in the numbers so we get it
abc- Guest
Ques 3
You measure the moment of inertia of a wheel by measuring its angular acceleration a = 0.413 rad/s2, when applying an external torque caused by a hanging weight of mass m = 200 g as shown above. The angular acceleration is decreased by a frictional torque, accounted for by afric = - 0.193 rad/s2 in the expression for the momemt of inertia I of the wheel, I = mr(g-ra) /(|afric|+a), where r is the radius of the cylinder attached to the wheel, and g is the acceleration of gravity, which you treat as error free. You neglect the error of the accelerations a and afric, and the error of the mass m. You find out that the term (a r) in the numerator is small compared to the term g, and thus neglect its error too. What is the absolute error of I, if the 2.5 cm radius r is known with a 1.3 % accuracy. (Hint: Read in the manual of the lab, how I can be written as two factors. One factor contains the quantities assumed to be error free, after taking into account that the term (a r) can be neglected for the purpose of error calculation. Use expression(3) and (4) in "Error and Uncertainty".) Indicate with a positive (negative) sign whether, with the neglections you made above, the absolute (relative) errors of I and r are the same.
Okay for those that wanted it step by step, using tofus method this is how to do it
delta r/r = %/100 * radius (in meters) so you do (1.3/100) * .025 m = .000325
put mass into kg for this next part, and use the positive afric since it asks for absolute value
I = mg/ (|afric|+angacceleration) = (.2 g * 9.81) / (.193 + .413) = (1.962)/(.606) = 3.2376
(.000325)*(3.2376) = answer = .00105222
And it is negative
hope that helps
Okay for those that wanted it step by step, using tofus method this is how to do it
delta r/r = %/100 * radius (in meters) so you do (1.3/100) * .025 m = .000325
put mass into kg for this next part, and use the positive afric since it asks for absolute value
I = mg/ (|afric|+angacceleration) = (.2 g * 9.81) / (.193 + .413) = (1.962)/(.606) = 3.2376
(.000325)*(3.2376) = answer = .00105222
And it is negative
hope that helps
Kathleen- Guest
Page 1 of 2 • 1, 2
Page 1 of 2
Permissions in this forum:
You cannot reply to topics in this forum
|
|