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Acceleration and KE of rotating cylinder

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Line joining the centre of path and cylinder is having angular velocity $\omega$ and angular acceleration $\alpha$ at the given instant:

In this
$V_{cm} = (R - r)\omega$

Total kinetic energy = $(1/2)mV_{cm}^2 + (1/2)I\omega ^2$
In this will we take moment of inertia (I) about point O or point C .

Also Acceleration of centre of cylinder = $(R-r)\alpha \hat{i} + (R-r)\omega ^2 \hat{j}$

But how could we find acceleartion of point of contact ?

asked May 12 in Physics Problems by koolman (4,116 points)
edited May 14 by sammy gerbil
The acceleration of the point of contact is $\alpha R$.  I do not understand your difficulty.
But in my book it is given as $\frac{\omega ^2 R(R-r)}{r} \hat{j}$
Ah you are asking about the point of contact on the moving cylinder, not on the surface it makes contact with
.
I think acceleration of Point of contact of surface would be zero as it is not moving .
There is no relative motion **tangentially** between the cylinder and the curved surface at the point of contact. But there is relative acceleration **radially**. A fixed point on the cylinder decelerates toward the surface then accelerates away again. See the cusps in the animation at https://en.wikipedia.org/wiki/Hypocycloid.
Yes , I agree with you . But I think we have to find at that instant .
Because the answers are given, you can use reasoning to decide which answer matches which question, without having to do any calculation.

Acceleration is a vector whereas KE is not. So answers 1, 2 will match with  R,S somehow, and answers 3, 4 with P, Q somehow.

The acceleration of point C definitely has an $\hat{i}$ component whereas we are not sure if the acceleration of P has an $\hat{i}$ component. So R matches 1 (the only answer with a $\hat{i}$ component) and by elimination S matches 2.

Total KE = rotational KE + translational KE. So the answer to Q (total KE) should be greater than P (rotational KE). Answer  4 > answer 3. Therefore P matches 3 and Q matches 4.

2 Answers

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Best answer

The following (much simpler) solution is that given by samjoe in your chat link :

$\vec{a_p} = \vec{a_c} + \vec{a_{p|c}}$

The cylinder has its own angular velocity $\Omega$ and angular acceleration $A$ about C. It is rolling without slipping so
$(R-r)\omega =r \Omega$
Differentiating
$(R-r)\alpha=rA$

The acceleration of the centre of cylinder C is
$\vec{a_{c}}= (R-r)\alpha \hat{i} + (R-r)\omega ^2 \hat{j}$
The acceleration of the point of contact P relative to C is
$\vec{a_{p|c}}=-rA\hat{i}+r\Omega^2 \hat{j}=-(R-r)\alpha \hat{i}+\frac{(R-r)^2}{r}\omega^2 \hat{j}$
Adding, the $\hat{i}$ components cancel out; only the $\hat{j}$ components remain.

Therefore
$\vec{a_p}=[(R-r)+\frac{(R-r)^2}{r}]\omega^2 \hat{j}=[1+\frac{R-r}{r}] (R-r)\omega^2 \hat{j}=\frac{R}{r}(R-r)\omega^2 \hat{j}$

answered May 14 by sammy gerbil (26,096 points)
selected May 14 by koolman
1 vote

Parametric equations for the trajectory of a point P on the rim of the cylinder are given in the wiki Hypocycloid by
$$x(\theta)=rk\cos\theta+r\cos(k\theta), y(\theta)=rk\sin\theta-r\sin(k\theta)$$ where $k=\frac{R-r}{r}$. From these we obtain an equation for the radial position of P : $$p(\theta)^2=x(\theta)^2+y(\theta)^2=r^2k^2+r^2+2kr^2\cos\theta\cos(k\theta)-2kr^2\sin\theta\sin(k\theta)=r^2(k^2+1)+2kr^2\cos (k+1)\theta$$
Next we differentiate twice wrt $t$ to find the radial velocity $\dot p$ and acceleration $\ddot p$ : $$2p\dot p=-2kr^2(k+1)\dot\theta\sin(k+1)\theta$$ $$\dot p^2+p\ddot p=-k(k+1)r^2[\ddot\theta\sin(k+1)\theta+(k+1)\dot\theta^2\cos(k+1)\theta] $$ We can substitute $\dot\theta=\omega, \ddot\theta=\alpha$. At the instant when point P reaches the rim we have $p=R, \dot p=0$ and for convenience we can set $\theta=0$. We then have $$R\ddot p=-k(k+1)^2r^2\omega^2$$ $$\ddot p=-\frac{R-r}{r}\frac{R^2}{r^2}\frac{r^2}{R}\omega^2=-\frac{R(R-r)}{r}\omega^2 $$


Comment: This is probably much more complicated than is necessary but I cannot see a simpler solution right now, and it does confirm that your answer is correct. The minus sign indicates that the radial acceleration is inwards because increases in $p$ are measured outwards.

answered May 13 by sammy gerbil (26,096 points)
edited May 14 by sammy gerbil
https://chat.stackexchange.com/transcript/message/44587937#44587937

While calculating Total K.E , we take moment of inertia (I) about which point O or point C ?
About whichever point is the axis of the rotation which you are considering. Moment of inertia depends on your choice of axis of rotation.
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