# 1 and 2

gueSt89 on Thu Oct 23, 2008 5:58 pm

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

super Mo on Thu Oct 23, 2008 7:02 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?

super Mo on Thu Oct 23, 2008 7:06 pm

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

dude on Thu Oct 23, 2008 7:42 pm

i am getting 4 wrong

any help?

guest23 on Thu Oct 23, 2008 8:01 pm

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

B on Thu Oct 23, 2008 8:25 pm

Not enough information for #1

We don't even have the radius

All we have is the masses lol

guest101 on Thu Oct 23, 2008 8:43 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)

Guest100 on Thu Oct 23, 2008 11:27 pm

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

J on Thu Oct 23, 2008 11:29 pm

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

anon on Fri Oct 24, 2008 12:35 am

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

Guest01 · Guest01 on Fri Oct 24, 2008 1:49 am

All the formulas are there given. They all worked for me. I think a few were mislabeled, but just read your question and you'll see which ones.

Question 1:
(Positive)
a=(m*g)/((M/2)+m)

Question 2:
(Negative)

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 1Cool 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

THANKS MIT!

To break it down just multiply across, watch order of operations!
L=((2/5)*M*R²)*(RPM*((2pi)/60))

Question 3:
(Negative)
w=(I'w')/I
becomes...
w=(1/I*w)
The ratio is given, but it becomes one over (1/x), multiply it by the w.

Question 4:
(Negative)
(M*%[NOTE: in decimal form]*9.81*L)+(m*9.81)]/(Sin(x)*.45[NOTE: should be the same number, but it's the second L given])

Hope that helps. See you all in a few hours!

super Mo on Fri Oct 24, 2008 5:58 am

I got the formula for #1 from the posts, but I can not figure out how. Can someone post the breakdown of the formulas. I would really like to know.

izzy on Fri Oct 24, 2008 3:16 pm

1.) answer is POSITIVE

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

where m=mass hanging and M=mass of the disk

2.) answer is NEGATIVE

(.4*m*(r^2))/(rpm*(2pi/60))

3.) answer is NEGATIVE

divide angular velocity w given by the value given for the ratio

4.) answer is NEGATIVE

((9.81*m)+(9.81*M*L1*(percent/100)))/(L2*(sin(theta)))

where m=weight held, M=mass of the person, L1=first L given, L2=second L given

hope this helps! any questions, get at me

Ch 8_3 help

#$

# 1 and 2

#1

Re: Ch 8_3 help

ques 3

Number 1

#3

Try rereading Number 1

#4

sucks =(

Re: Ch 8_3 help

#1

simplified solution to all 4 questions