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Block and cylinder down an incline

by vladimir69
Tags: block, cylinder, incline
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vladimir69
#1
Mar20-09, 02:26 AM
P: 131
1. The problem statement, all variables and given/known data
A block of mass m1 is attached to the axle of a uniform solid cylinder of mass m2 and radius R by massless strings. The two accelerate down a slope that makes an angle [itex]\theta[/itex] with the horizontal. The cylinder rolls without slipping and the block slides with coefficient of kinetic friction [itex]\mu[/itex] between the block and slope. The strings are attached to the cylinders axle with frictionless loops so that the cylinder can roll freely without any torque from the string. Find an expression for the acceleration of the pair, assuming that the string remains taut.


2. Relevant equations
F=ma
T=tension in string
N=normal force
a=acceleration of system
3. The attempt at a solution
I applied Newton to the block and cylinder separately.
For the block (choosing axis with x direction parallel to the slope and y direction perpendicular to the slope)
y direction:
[tex]N-m_{1}g\cos\theta=0[/tex]
x direction:
[tex]T-\mu N + m_{1}g\sin\theta=m_{1}a[/tex]

For cylinder
x direction:
[tex]-T+m_{2}g\sin\theta = m_{2}a[/tex]

I can solve for the acceleration but its not in agreement with the answer in the book, so perhaps I am leaving something out. Could be the torque on the cylinder but not sure about that.

Anyone got any ideas?
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sArGe99
#2
Mar20-09, 02:46 AM
P: 133
Did they give a diagram along with the question?
Which moves first? The block or the cylinder?
and if you meant 'strings' then there should be tension from each string acting on the block, right?
vladimir69
#3
Mar20-09, 05:13 AM
P: 131
yeah there was a diagram. the cylinder is closer to the bottom and the block is up the top

sArGe99
#4
Mar20-09, 05:28 AM
P: 133
Block and cylinder down an incline

Quote Quote by vladimir69 View Post
yeah there was a diagram. the cylinder is closer to the bottom and the block is up the top
Yea, thanks for that info. Let me see. hm
sArGe99
#5
Mar20-09, 05:29 AM
P: 133
There are two strings connecting the objects, right?
sArGe99
#6
Mar20-09, 05:40 AM
P: 133
I got an answer now and would like to know if its correct.
[tex]a = 2 \frac{(m_{1} + m_{2})g sin\theta - \mu m_{1}g cos\theta}{3 m_{2} + 2 m_{1}}[/tex]
vladimir69
#7
Mar20-09, 08:21 AM
P: 131
yeah that looks about right
how did you go about getting that result
sArGe99
#8
Mar21-09, 01:15 AM
P: 133
Write the correct force equations for both the block and the cylinder and also the torque equation for the cylinder since its rolling.
Hint : Friction provides torque for the cylinder to rotate.
vladimir69
#9
Mar21-09, 02:49 AM
P: 131
ahh yes the correct force equations... i am guessing i have them correct.
i am not sure about the torque on the cylinder, perhaps gravity also provides torque to the cylinder?
but something must be wrong because according to my equations i can solve for the acceleration without the need for the torque equation
sArGe99
#10
Mar21-09, 04:43 AM
P: 133
Gravity is a central force. It can't quite provide torque for rotation. Only friction does it in this case.
vladimir69
#11
Mar21-09, 06:01 AM
P: 131
note: x axis taken to be parallel to the slope of the incline
Torque equation for cylinder:
[tex]-Rm_{2}g\sin\theta=\frac{1}{2}m_{2}R^2\alpha[/tex]

=>[tex]R=-\frac{2g}{\alpha}\sin\theta[/tex]

Forces for block: y direction
[tex]N_{1}-m_{1}g\cos\theta=0[/tex]

Forces on block: x direction (*)
[tex]
T-\mu N + m_{1}g\sin\theta=m_{1}a
[/tex]

Forces on cylinder: x direction (*)
[tex]
-T+m_{2}g\sin\theta - \mu N_{2} = m_{2}a
[/tex]

Forces on cylinder: y direction
[tex]N_{2}-m_{2}g\cos\theta=0[/tex]

The only difference with my first answer is that I added friction to the force equation for the cylinder along the x axis, but i still dont need the torque equation to solve for acceleration. So I think it can be narrowed down to either block or cylinder force equation along the x axis (marked with asterisks). The only thing I can think of is the accelerations, a, for those equations is different but that doesnt sound right to me as they should both be accelerating at the same rate down the hill.
Doc Al
#12
Mar21-09, 06:13 AM
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Doc Al's Avatar
P: 41,476
Careful: You cannot assume that the friction force on the cylinder equals μN. That friction force is static friction, and it will have whatever value is required to prevent slipping up to a maximum of μN. Hint: Call the friction force on the cylinder F and treat it as one of your unknowns.

You have three unknowns and you'll need three equations.
vladimir69
#13
Mar21-09, 07:57 AM
P: 131
note: x axis taken to be parallel to the slope of the incline
Torque equation for cylinder:
[tex] -RF_{f}=\frac{1}{2}m_{2}R^2\alpha[/tex]
=>
[tex]F_{f} = -\frac{1}{2}m_{2}R\alpha [/tex]

Forces for block: y direction
[tex]N-m_{1}g\cos\theta=0[/tex]

Forces on block: x direction (*)
[tex]T-\mu N + m_{1}g\sin\theta=m_{1}a[/tex]
[tex]T=\mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a[/tex]

Forces on cylinder: x direction (*)
[tex] -T+m_{2}g\sin\theta - F_{f} = m_{2}a[/tex]
[tex]T= m_{2}g\sin\theta - F_{f} -m_{2}a[/tex]

popping everything in
[tex] \mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a=m_{2}g\sin\theta +\frac{1}{2}m_{2}R\alpha -m_{2}a [/tex]
but
[tex]a=R\alpha [/tex]
so
[tex] \mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a=m_{2}g\sin\theta +\frac{1}{2}m_{2}a -m_{2}a [/tex]
[tex](m_{1}+m_{2})g\sin\theta-\frac{1}{2}m_{2}a=m_{1}a+\mu m_{1} g \cos\theta[/tex]
[tex]m_{1}a+\frac{1}{2}m_{2}a = (m_{1}+m_{2})g\sin\theta+\mu m_{1} g\cos\theta[/tex]
[tex]a(m_{1}+\frac{1}{2}m_{2})=(m_{1}+m_{2})g\sin\theta+\mu m_{1} g\cos\theta[/tex]
something like that anyway
i am more interested in knowing did i get the 3 equations for the 3 unknowns right?

thanks for the help Doc Al
and also sarge99
sArGe99
#14
Mar21-09, 08:01 AM
P: 133
So that is the answer. One of my force equations was wrong :O
Thanks for the answer, vladimir
Doc Al
#15
Mar21-09, 08:11 AM
Mentor
Doc Al's Avatar
P: 41,476
Quote Quote by vladimir69 View Post
note: x axis taken to be parallel to the slope of the incline
Torque equation for cylinder:
[tex] -RF_{f}=\frac{1}{2}m_{2}R^2\alpha[/tex]
=>
[tex]F_{f} = -\frac{1}{2}m_{2}R\alpha [/tex]
Get rid of that minus sign. Express alpha in terms of a, which you do later. This is your first equation.

Forces for block: y direction
[tex]N-m_{1}g\cos\theta=0[/tex]

Forces on block: x direction (*)
[tex]T-\mu N + m_{1}g\sin\theta=m_{1}a[/tex]
[tex]T=\mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a[/tex]
Good. This is your second equation.

Forces on cylinder: x direction (*)
[tex] -T+m_{2}g\sin\theta - F_{f} = m_{2}a[/tex]
[tex]T= m_{2}g\sin\theta - F_{f} -m_{2}a[/tex]
Good. This is your third equation.

popping everything in
[tex] \mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a=m_{2}g\sin\theta +\frac{1}{2}m_{2}R\alpha -m_{2}a [/tex]
but
[tex]a=R\alpha [/tex]
so
[tex] \mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a=m_{2}g\sin\theta +\frac{1}{2}m_{2}a -m_{2}a [/tex]
[tex](m_{1}+m_{2})g\sin\theta-\frac{1}{2}m_{2}a=m_{1}a+\mu m_{1} g \cos\theta[/tex]
[tex]m_{1}a+\frac{1}{2}m_{2}a = (m_{1}+m_{2})g\sin\theta+\mu m_{1} g\cos\theta[/tex]
[tex]a(m_{1}+\frac{1}{2}m_{2})=(m_{1}+m_{2})g\sin\theta+\mu m_{1} g\cos\theta[/tex]
something like that anyway
You have a sign error, carried from your first equation above. (But almost there! )
vladimir69
#16
Mar21-09, 09:41 PM
P: 131
Got it thanks Doc Al


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