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Time for Cylinder to Travel Roll Down Incline

  • Thread starter seichan
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Homework Statement
A cylinder (R= 0.11 m, I (center of mass)= 0.015427 kg*m2, and M= 1.48 kg) starts from rest and rolls without slipping down a plane with an angle of inclination of Theta= 26.3 deg. Find the time it takes it to travel 1.51 m along the incline.


The attempt at a solution
Alright, so I tried to deal with this one using torque. The torque is due to the plane's reaction to the weight of the cylinder, so...
G=m*g*r*sin(theta)=1.48*9.81*.11*sin(26.3)=.707614820587 N
G=I*a
a=G/I
a=.708/.015847=44.6529198326 rad/s^2

Now, calculate how far in radians the distance traveled is...
(2pi*1.51)/.11= 86.2509983076 rad

So, using the position equation...
s=.5at^2
t=sqrt(2s/a)
=sqrt(2*86.2509983076/44.6529198326)
=1.96549585305 s

Yeah... That's wrong... Any clue where I'm wrong? Thanks for any help you guys provide- you're all really great =)
 
Last edited:

Answers and Replies

Dick
Science Advisor
Homework Helper
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The distance travelled in radians is 2*pi*d/r, isn't it? Isn't that 2*pi*1.51m/0.11m? How did the mass get into that equation?
 
32
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Oops- yeah, that was a mistake I made when I was typing that into here. The other way still comes out wrong.
 
454
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if the cylinder isn't moving there is a friction force mgsin(theta) at a distance r that will produce a torque, but if the cylinder has a linear accelaration a, the magnitude of this force is mgsin(theta) - ma
 
412
2
The error in your analysis is that you are considering that the normal force causes torque. The fact is, the gravitational force has two components. The normal force is the reaction to one of them i.e. mg cos(θ).

However, all the three forces have their lever arm as 0. [the gravitational forces act at the center and hence r = 0, and the normal force is parallel to the radial line and hence θ = 0]. This diagram may help you:

http://img369.imageshack.us/img369/5736/torquefbdqd0.jpg [Broken]

When the cylinder is acted upon by the gravitational force component, mg sin(θ), it starts to move. Just as it starts to move, frictional force acts upon it. This force has a lever arm 'r' and this is what causes the torque. Since this is a case of pure rolling, you can use the formulas [itex]\alpha = ar[/itex], [itex]v = \omega r[/itex] and [itex]s = \theta r[/itex] without hesitation.

This is also a case of constant acceleration [both, linear and angular]. One suggestion i could give here is to use conservation of energy. What is the GP energy of the cylinder at a particular height? How much does this decrease when it rolls down? This difference of energy will be provided by a change in R.KE and L.KE. Using those formulae, you can compute the angular velocity when the cylinder has rolled down. Once you do that, simply use the formulae for constant angular acceleration to find out the time.

P.S: Try doing it without using any of the values provided, but putting in only variables in their place. You'll find that the time taken depends only on the shape, inclination of the plane and distance travelled i.e. mass doesn't come into the equation at all :D
 
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