How Do You Calculate Motion Parameters of a Rolling Object on an Incline?

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To calculate the motion parameters of a roller on an incline, it is essential to understand that the roller accelerates uniformly from rest, meaning its velocity changes throughout its descent. The time taken for each rotation is not constant due to this acceleration, contradicting the assumption that each revolution takes the same time. The correct approach involves using kinematic equations for rotational motion, as the roller's initial velocity is zero and its acceleration is not zero. The formula v = 2πr/t is incorrect in this context because it implies constant velocity, which does not apply here. To solve the problem accurately, one must apply the appropriate kinematic equations for rotational motion under constant acceleration.
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5. A roller 0.44 m diameter rolls down a slope starting from rest. It takes 12 seconds make 6 complete rotations along the sloping surface accelerating uniformly as it moves. Calculate the following:

(1). The angular velocity at the end.

(2). The linear velocity at the end.

(3). The angular acceleration.

(4). The linear acceleration.

(5). The distance travelled.




from the question its obvious each rotation takes two seconds

12/6=2
then i thought omega= (2*pi)/12 gives you rads per second
but i then got really confused as i actualy don't have the velocity in the question and i lost it all

is the angular velocity (number of revs*2*pu)/number of cycle?

any ideas? i am losttt!
 
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Start from the kinematic equations for rotational motion. Although 12/6 is always equal to 2, this is irrelevant here. Each complete revolution takes less time as the roller accelerates down the incline. Furthermore, the angular velocity also changes as time increases.
 
but wouldn't all reveloutions be the same since the rollers speed is the same and in move uniformly?
 
also is this equation right to work out the final velocity?

v=2*pi*r/t

then the answer times the number of rotation?
or more needs to be done or is this wrong?
 
aghakarim said:
but wouldn't all reveloutions be the same since the rollers speed is the same and in move uniformly?
The roller's speed is not the same at all times. It starts from rest and this means that its initial velocity is zero. If the roller's velocity were the same at all times, its velocity would remain at zero and the roller would stay where it is, at the top of the incline. This is not what happens. The roller rolls down the incline which means that its velocity changes from zero to something other than zero. Therefore the roller accelerates and this means that its velocity is not the same at all times.
also is this equation right to work out the final velocity?

v=2*pi*r/t
It is not right because it assumes that the acceleration is zero, i.e. that the roller makes the same number of revolutions in the same amount of time. As I just pointed out, the acceleration cannot be zero.

Like I said, what are the kinematic equations for rotational motion under constant acceleration? There are four of them. Write them down and see whether you can figure out which one to use considering what is given to you.
 
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