Wheel rolling on a horizontal surface

[Jahnavi]

@kuruman , I have replied your question in post#53 . Would like to read your views :)

 

[kuruman]

The point on the ground does not behave as the instantaneous point of rotation for all intents and purposes. As the OP has discovered, it behaves as such in terms of velocity but not in terms of acceleration.

Yes, of course. Bad choice of words.

The suggestion by @collinsmark in post #50 is simple and works well. One can also write an equation giving the position vector of point P on the wheel as a function of time, take the double derivative and evaluate the acceleration at a time when the position vector is vertical. The results are the same.

 

[Jahnavi]

Yes, you could still choose point O as the center of rotation. But in your final answer you would have to add the acceleration of point O itself because point O itself is accelerating.

@PeroK ,

Tangential acceleration of point P w.r.t O is 4m/s² .

Radial acceleration of P w.r.t O is 12m/s² (vertically down ) .

Acceleration of O in the lab frame is 9 m/s² ( vertically up ).

Net acceleration of point P in the lab frame would be vector sum of the above three components .

Is that correct ?

 

[PeroK]

@PeroK ,

The tangential acceleration of point P w.r.t O is 4m/s² .

Radial acceleration of P w.r.t O is 12m/s² (vertically down ) .

Acceleration of O in the lab frame is 9 m/s² ( vertically up ).

Net acceleration of point P in the frame of O would be vector sum of the above three components .

Is that correct ?

My suggestion for this problem is to test any solution against the known acceleration of the centre of the wheel.

As long as you are happy with how you got the acceleration of O, then yes that's correct.

The final answer is in the lab frame, of course.

 

[Jahnavi]

As long as you are happy with how you got the acceleration of O, then yes that's correct.

OK.

So all I had to do was to add the acceleration of O vectorially to the accelerations in post#7 so as to get acceleration of P in lab frame .

The answer also matches .

 

[collinsmark]

OK.

So all I had to do was to add the acceleration of O vectorially to the accelerations in post#7 so as to get acceleration of P in lab frame .

The answer also matches .

Yes, that's essentially correct. In post #7 you neglected to convert back to an inertial frame. Once you account for the acceleration of frame O, you'll get the correct answer.

If you haven't done so yet, you owe it to yourself to try the problem again but this time choose the wheel's center as your accelerating frame of reference. The process is the same, i.e., you'll need to add (vector wise) the acceleration of your accelerating frame (the wheel's center in this case) to your preliminary result before you are finished, in the process of converting back to the inertial frame.

So either way gives you the right answer. But now, since you've done it both ways, you can compare which method is the more straightforward of the two.