# Why does $a=R\alpha$ hold here? A cylinder is suspended by two strings wrapped around it. The strings gradually unwind and the cylinder moves down with acceleration $a$ and angular acceleration $\alpha$.

Why does my textbook say that $a=R\alpha$? I could not think of any reason why the equality should hold. Can someone please explain?

• I am sure if they have written the equality holds then they must also have given that there is no slipping between strings and the cylinder. If they haven't, then there must be some printing error. Jan 5, 2017 at 13:13
• @SagarKaushik Yes, you can consider that there is no slipping. But still I don't understand why the equality holds.
– user102705
Jan 5, 2017 at 13:14
• @tomph I am asking why the linear acceleration of the cylinder is equal to the R times the angular acceleration.
– user102705
Jan 5, 2017 at 13:19

Because the string unwinds. Thus, in order to move down a distance $x$, the cylinder has to rotate by the angle $\phi = \frac{x}{R}$.

The accelerations are $a = \ddot{x}$ and $\alpha = \ddot{\phi}$, respectively. Thus $a = R\alpha$.

Said by other words, think of what the equation $a=R\alpha$ represents: $a$ is the acceleration of a particle on the rim (a distance $R$) from the centre).

On a simple freely spinning wheel, the rim is moving with $a$. On a wheel of a car, the rim is actually standing still at the point of contact with the road. Here instead the centre is moving with $a$. And the opposite rim-point moves with $2a$.

The point is that this $a$ is the motion (acceleration) of a rim-point relative to the centre.

Back to your situation. Your strings are hanging still. No acceleration after unwinding. So the final touching point where the strings touch the cylinder is standing still in that moment, just like the contact-point of wheel-on-road for a car. This is the same situation as the car's wheel, where it is the centre instead of this rim-point that accelerates with $a$. And the opposite rim-point has $2a$.

The rim-point moves with $a$ seen from the centre-point and because of that the relation $a=R\alpha$ holds.

• Loved the example! Jan 5, 2017 at 14:12

Take a point on the circumference where the string touches the cylinder as fixed.

Now unwrap the string a little bit $\Delta \theta$ more. If the string is hanging vertically, the center of the cylinder will move down by $r\Delta \theta$. Differentiate once to get velocity, and twice to get acceleration. So if you denote angular acceleration by $\alpha$, the result follows.

• That is okay. No probs. I realized that if the cylinder moves down by $x$ then I need to find the angle $\theta$ which makes that length on the circumference. And that is why $x=r\theta$. On differentiating it I get the relation. Thanks a lot for your help btw :)
– user102705
Jan 5, 2017 at 13:28

The angular acceleration $\alpha$ times the radius $r$ of the cylinder is the acceleration on the circumference of the cylinder $a$. (it's $c=2\pi r$)

• The motion of the cylinder downward in the rest frame, and the string upward in the cylinder frame are equal and opposite Jan 5, 2017 at 13:23

At the point where string just leaves the contact of the cylinder (string is tangent at that point to the cylinder) the downward acceleration is 'a' (translational acceleration of the centre of mass) and the "upwards acceleration because of rotation" at that point is 'R$\alpha$' and they must be equal so that the velocity of that point is always 0. It should be = 0 because if not, then there will be relative slipping between the string and the cylinder as string has velocity = 0 (the string is attached to the roof which is obviously at rest and therefore it must have 0 velocity or else it will break).