# Rolling Without Slipping

1. Jun 15, 2009

### arpan251089

A cylinder rolls up an inclined plane, reaches some height, and then rolls down. The directions of frictional force acting on the cylinder while descending will be up the inclined. But what will be the direction of frictional force when the cylinder is ascending??
UP or DOWN

Well, I think the answer must be down the inclined. But the book says it will be up.
The reason given in the book is
mgsinθ acts in downward direction in both cases. So the sense of angular acceleration should also be same in both cases. So friction always acts upwards

Pls help if anyone understands the reason.

2. Jun 15, 2009

### fatra2

Hi there,

Very simply put, friction always acts against the movement. Therefore, if the object moves upward, then friction will be acting downwards. Secondly, friction, in classical physics, will always be parallel to the plane of movement.

Cheers

3. Jun 15, 2009

### Starwatcher16

Are you sure its not talking about the normal force?

As for friction, I am not sure why its asking for it. As there is no work done by it (in this idealied case).

4. Jun 15, 2009

### tiny-tim

Hi arpan251089!
Sorry, but nobody is reading this question right.

This is a question about circular motion and force, not linear.

If there was no friction (if the slope was ice), then the axis of the cylinder would accelerate down the slope (or decelerate up it), but it would keep the same angular velocity.

So you should ask yourself:

i] what is the angular acceleration of the cylinder (either going up or down)

ii] how many forces are there producing a torque which can affect this?

5. Jun 15, 2009

### arpan251089

I agree you tiny-tim and fatra2

@fatra2
Yes you are right that friction acts opposing the velocity and not applied forces. But here we are talking abt angular motion and not linear motion. This cannot be applied in rolling motion. See the reason below.

@tiny-tim
Thanks. After thinking for sometime i came to a conclusion.

See friction acts in upward direction while descending. This is 100% correct. Due to this a torque acts on the cylinder such that it rolls down with INCREASING angular velocity.
But when the cylinder is ascending, it will be supplied with kinetic energy[Rotational] which will become zero at the top. Thus angular velocity will decrease. This means friction will try to exert torque in a direction to oppose the angular velocity. And for this the friction again will have to act in upward direction

6. Jun 15, 2009

### rcgldr

Assuming this is a solid cylinder of uniform density,

given:

a = acceleration
af = acceleration component due to friction force
ag = acceleration component due to gravity
f = friction
r = radius of cylinder
t = torque
θ = angle of plane
ω = angular speed
α = angular acceleration
i = angular inertia

a = α / r

for a solid cylinder, i = 1/2 m r2

for the downhill rolling case:

m a = m g sin(θ) - f

t = f r = i α = 1/2 m r2 (a / r)

f = t / r = 1/2 m a

m a = m g sin(θ) - 1/2 m a
a = 2/3 g sin(θ)

f = 1/3 m g sin(θ)

ag = g sin(θ)
af = 1/3 g sin(θ)
a = ag - af = 2/3 g sin(θ)

For the uphill case, the linear and angualr velocities differ, but not the linear and angular accelerations, so even in the uphill case, the angular deceleration generates a downhill force on to the surface of the plane, opposed by an uphill friction force.

In this case, the uphill af remains = 1/3 g sin(θ), opposed by the component of gravity paralled to the plane = g sin(θ), and the linear rate of acceleration remains = 2/3 g sin(θ).

In both cases, the friction force is uphill, = 1/3 m g sin(θ), extending the distance rolled uphill because of the relationship to angular inertia, and decreasing the rate of downhill acceleration.

other cases:

For a solid sphere, i = 2/5 m r2, af = 2/7 g sin(θ)
For a solid cylinder, i = 1/2 m r2, af = 1/3 g sin(θ)
For a hollow sphere, i = 2/3 m r2, af = 2/5 g sin(θ)
For a hollow cylinder, i = m r2, af = 1/2 g sin(θ)

Last edited: Jun 15, 2009
7. Jun 16, 2009

### tiny-tim

Hi arpan251089!
Yes, that's right , but you can put it more simply …

linear acceleration (of the c.o.m.) is always downhill,

so the angular acceleration must also always be "downhill" (obviously, that needs defining a little better ),

and the only torque about the c.o.m. comes from the friction, so that must be uphill.