Ch 8_3 help

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Ch 8_3 help

Post  Specs on Tue Oct 21, 2008 12:44 am

Can't seem to get these two problems, any help would be greatly appreciated....

3) A mass m = 0.7977 kg hangs from a 3.593 kg uniform disk via a string which is wound around the circumference of the disk (see sheet 21 with no counter weight on the left hand side). After you release the mass it accelerates downward. (Assume that the the string does not slip while unwinding from the disk). What is the linear acceleration of the mass ? (see sheet 23 on how to express the torque due to m in terms of the linear acceleration, then relate this net torque to the angular acceleration as on sheet 22; relate the angular acceleration to the linear one and solve for it; the moment of inertia is given on sheet 18; note that the radius of the disk drops out in the final expression for the linear acceleration)


6) Use the model for a 85.11 kg person bending over and holding a weight while the torso is in a horizontal position given on sheet 42,45'. The weight held has a 15.9 kg mass and has a horizontal distance L from the hip joint (the pivot point). The torso, head and arms taken together (modeled by the ruler) is 65% of the body weight with a center of mass point at a distance of about 0.609L away from the hip. Assume that the point of action for the various back muscles holding the torso in the horizontal position (modeled by the cable) is a distance 0.45L from the hip. What is the tension of "the" backmuscle which makes a 120 degree angle with the horizontal (see sheet 43,44 top) ? Indicate with a negative (positive) sign whether the tension is larger (smaller) than the body weight.

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Ch 8.3 Help

Post  sedwards on Tue Oct 21, 2008 5:36 pm

Question 1: (1 point)

A mass m = 0.645 kg hangs from a 1.93 kg uniform disk via a string which is wound around the circumference of the disk (see sheet 21 with no counter weight on the left hand side). After you release the mass it accelerates downward. (Assume that the the string does not slip while unwinding from the disk). What is the linear acceleration of the mass? (See sheet 23 on how to express the torque due to m in terms of the linear acceleration, then relate this net torque to the angular acceleration as on sheet 22; relate the angular acceleration to the linear one and solve for it; the moment of inertia is given on sheet 18; note that the radius of the disk drops out in the final expression for the linear acceleration.)

(how to solve):

Question 2: (1 point)

A uniform 2.899 kg sphere (the factor f=2/5 on sheet 18) with a 38.87 cm radius spins at 47.21 rpm. What is the angular momentum of the sphere (see sheet 26). Indicate with a negative (positive) sign whether the angular momentum stays the same (changes) in the absence of any torque acting on the sphere.

(how to solve):

Question 3: (1 point)

A diver makes a series of summersaults during a diving competition. Stretched out he rotates about an axis which is horizontal and perpendicular to his body. He then goes into a tuck reducing his moment of inertia by a factor of 5.013 and making now 1.0801 revolutions per second. What was his initial angular velocity in revolutions per second (see sheet 27)? Indicate with a negative (positive) sign whether inclusion of air resistance and thus inclusion of an external torque would (not) modify your answer.

(how to solve):

Question 4: (1 point)

Use the model for a 80.19 kg person bending over and holding a weight while the torso is in a horizontal position given on sheet 42,45'. The weight held has a 21.89 kg mass and has a horizontal distance L from the hip joint (the pivot point). The torso, head and arms taken together (modeled by the ruler) is 65% of the body weight with a center of mass point at a distance of about 0.6254 L away from the hip. Assume that the point of action for the various back muscles holding the torso in the horizontal position (modeled by the cable) is a distance 0.45L from the hip. What is the tension of "the" backmuscle which makes a 120 degree angle with the horizontal (see sheet 43,44 top)? Indicate with a negative (positive) sign whether the tension is larger (smaller) than the body weight.

(how to solve):

sedwards
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quiz questions

Post  angelbab on Tue Oct 21, 2008 5:37 pm

Could someone please help with these I am so confused?

angelbab
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Re: Ch 8_3 help

Post  Specs on Tue Oct 21, 2008 11:25 pm

In reply to sedwards

Question 4 is simply:

[(80.19*.65*9.81*.6254)+(21.89*9.81)]/(Sin(120)*.45)

Still need help with number one though…any help would be great....

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Re: Ch 8_3 help

Post  Specs on Tue Oct 21, 2008 11:48 pm

Ok for Qu 1 I believe this should work out...

a=(m*g)/((M/2)+m)

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Question 2 and 3

Post  hwilson on Wed Oct 22, 2008 12:51 am

Could somone please help me with question 2 and 3?

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Question 2 and 3

Post  guest22 on Wed Oct 22, 2008 10:26 am

I need help with question 2 and 3 does anyone have any clue?

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signs

Post  kat on Wed Oct 22, 2008 6:50 pm

anyone know the signs for each of the answers? whether they're pos or neg?

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Re: Ch 8_3 help

Post  Guesto on Wed Oct 22, 2008 9:24 pm

can somebody post an explanation for this shit... numbers and formulas dont really help at all...

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#2

Post  Mit on Wed Oct 22, 2008 11:24 pm

Here's the method:
L (angular momentum) = I (moment of inertia) * W (angular velocity).
therefore, I = (2/5) factor that was provided in the question * M (in Kg)* (R^2), R being in meters! and W = rpm given in the question * ((2*pi)/60) = (rads/seconds)
Multiply the I and W and you get your angular momentum:
Here is my problem and answer:
A uniform 2.728 kg sphere (the factor f=2/5 on sheet 18) with a 53.37 cm radius spins at 33.42 rpm. What is the angular momentum of the sphere (see sheet 26). Indicate with a negative (positive) sign whether the angular momentum stays the same (changes) in the absence of any torque acting on the sphere.
Ans: -1.087761858

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Question 3

Post  james on Thu Oct 23, 2008 1:50 pm

Anyone any ideas on number 3?

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Re: Ch 8_3 help

Post  guest11 on Thu Oct 23, 2008 3:42 pm

could somebody please post the signs for the problems

thanks

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#4

Post  PHY999 on Thu Oct 23, 2008 4:41 pm

I'm having difficulty using the method proposed above for question 4. Can someone please clarify and include whether or not its pos or neg??

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#3

Post  guest88 on Thu Oct 23, 2008 6:12 pm

for 3 the answer is simple. sheet 27 states Iw=I'w', so just solve for w which is the initial...rearrange the formula and you get w=(I'w')/I....note I is just one and I' is 1/factor given...and the answer is negative

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signs for 8_3

Post  an on Thu Oct 23, 2008 6:31 pm

1) +
2) -
3) -
4) -

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#$

Post  gueSt89 on Thu Oct 23, 2008 6:58 pm

any1 know how to do #4 that solution given does not work....thats the only one im stuck on

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# 1 and 2

Post  super Mo on Thu Oct 23, 2008 8:02 pm

# 1: I = m*r^2
#2 formula:
(m_1 * [%/100]*9.81*L_1+m_2*9.81)/(sintheta*L_2)
How does that work for 'yall?

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#1

Post  super Mo on Thu Oct 23, 2008 8:06 pm

Forgett the formula for #1 I was looking at an old homework. So sorry, #4 should be right though.

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Re: Ch 8_3 help

Post  dude on Thu Oct 23, 2008 8:42 pm

i am getting 4 wrong

any help?

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ques 3

Post  guest23 on Thu Oct 23, 2008 9:01 pm

do we change revs per s to rads/s???

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Number 1

Post  B on Thu Oct 23, 2008 9:25 pm

Not enough information for #1

We don't even have the radius

All we have is the masses lol

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#3

Post  guest101 on Thu Oct 23, 2008 9:43 pm

for #3 you dont need to convert rev/s to rad/s...just leave it as is and use the equation that was given...

w=(I'w')/I
(I' being 1/factor..w' being rev/s #...and I being 1)

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Try rereading Number 1

Post  Guest100 on Fri Oct 24, 2008 12:27 am

The last sentence... "note that the radius of the disk drops out in the final expression for the linear acceleration"

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#4

Post  J on Fri Oct 24, 2008 12:29 am

(M*(%/100)*g*L1)+(m*g))/(sintheta*L2)

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sucks =(

Post  anon on Fri Oct 24, 2008 1:35 am

this sucks. where is the person who posted the step by step solutions for the other ones? =(

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Re: Ch 8_3 help

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