Slipping/Rolling Motion on Inclined Plane

[michelp]

TL;DR: I may have some problems with sign of convention and stuff

Attempt at solution and correct answer is attached. the linear acceleration indicates that the friction actually is in the same direction as linear speed.

Why is the angular acceleration anticlockwise here, against the direction of angular speed? Shouldn't angular speed increase? Please help

 

[Lnewqban]

Please, note that the value of the final angular velocity is 20t=24 rad/s.
The dynamic friction of the rotating-sliding cylinder with the rough inclined plane consumed part of the original rotational energy.

 

[michelp]

Please, note that the value of the final angular velocity is 20t=24 rad/s.
The dynamic friction of the rotating-sliding cylinder with the rough inclined plane consumed part of the original rotational energy.

I'm sorry, could you expand more on that?

 

[PeroK]

Why is the angular acceleration anticlockwise here, against the direction of angular speed? Shouldn't angular speed increase? Please help

The cylinder is initially spinning in a clockwise direction. The frictional force on the cylinder is down the incline. This provides a linear acceleration and a rotational deceleration.

 

[michelp]

The cylinder is initially spinning in a clockwise direction. The frictional force on the cylinder is down the incline. This provides a linear acceleration and a rotational deceleration.

so if I'm getting it right, it depends on the initial condition, right? if the cylinder at first only does translation, then the friction would be against the speed of center of mass, right?

 

[PeroK]

so if I'm getting it right, it depends on the initial condition, right? if the cylinder at first only does translation, then the friction would be against the speed of center of mass, right?

Yes. In that case, there will be linear deceleration and rotational acceleration until the state of rolling without slipping is reached.

 

[A.T.]

if the cylinder at first only does translation, then the friction would be against the speed of center of mass, right?

Yes, the direction of friction is opposite to the relative motion at the contact point. And that direction depends on: linear velocity of the center, angular velocity and radius.

 

[Steve4Physics]

if the cylinder at first only does translation

I guess you meant to say something like "if the cylinder's initial angular speed = 0 and it then rolls without slipping".

then the friction would be against the speed of center of mass, right?

Not quite right.

For rolling-without slipping (RWS), the direction of the frictional force is always uphill. (You might want to think about why this is so!)

So for RWS uphill, the friction is in the same direction as the centre-of-mass velocity.

And for RWS downhill, the friction is in the opposite direction to the centre-of-mass velocity.

Also,you used the phrase:

... against the speed of center of mass

Note that 'speed' is not a vector so it has no direction. You probably meant to say something like " in the direction opposite to the velocity of the centre of mass".

 

[Lnewqban]

I'm sorry, could you expand more on that?

Just tried to answer your questions:
Why is the angular acceleration anticlockwise here, against the direction of angular speed?
Because the rotation of the cylinder is slowing down with time.

Shouldn't angular speed increase?
It should, if its value at release time was zero, and friction was present.
In this case, the cylinder is said to be spinning at release time faster than it would if naturally rolling downhill.

What happened to that initial rotational energy?