# What would be the acceleration of a ball moving in uniform circular motion?

Question in a book:

A ball of $mass = 0.5kg$ is attached to the end of an inelastic string of length $L=0.5m$. The other end of the string is fixed. The ball is made to move on a horizontal circular path about the vertical axis through the fixed end of the string. The maximum tension that the string can bear is 324 N. The maximum possible angular velocity of the ball is 36 rad/s, the acceleration of the ball along the string is?.

Solution given:

Since the string is inelastic acceleration along the string is zero.

My problem with the solution:

Why on the earth would it be zero. If the ball is rotating with $\omega$ then shouldn't the centripetal acceleration which is along the string directed towards the center be $r\omega^2$? But it isn't mentioned in the problem with what $\omega$ the string is rotating. Then how to go about solving the problem?

PS: I am a physics tutor for 11th and 12th grades.

• "the acceleration of the ball along the string is zero" This is simply impossible if that ball makes a circular path. They even mention the tension in the string, which whould point towards acceleration along the string. Which physics book is this? Feb 16, 2017 at 10:49
• It is part of an exercise in a handout given in class. Feb 16, 2017 at 10:51
• @Steeven The centripetal force is acting along the string? Feb 16, 2017 at 10:56
• I guess what they mean by this is that the string does not change length. Or said differently, if you take a rotating reference frame, the ball does not move. But I agree that at least the wording of the question is rather confusing. Feb 16, 2017 at 11:19
• @user1583209 In a rotating reference frame, I agree with you. But should I get same answer from both inertial and non-inertial frame? Feb 16, 2017 at 11:25

The question is very poorly worded and confusing.

The fact that the ball is forced to move in a circular path means that the radial velocity is zero but the radial acceleration is $a_r=r\omega^2$ as you correctly point out. Sufficient information is provided to calculate this acceleration, assuming that the angular velocity $\omega$ and tension $T$ are the maximum possible.

The fact that the string is inextensible is irrelevant, because a ball at the end of an elastic string can also be forced to move in a perfect circle.

What is described in the question is a conical pendulum. The string does not lie along the radius of the circle which the ball moves in. The radius of this circle is $r=L\sin\theta$ where $\theta$ is the inclination of the string to the vertical. The tension $T$ in the string is such that
$T\cos\theta=mg$
$T\sin\theta=ma_r$
where $a_r$ is the centripetal acceleration. From these you can eliminate $\theta$ to get
$a_r^2=(\frac{T}{m})^2-g^2$.

There is a non-zero component of centripetal acceleration along the direction of the string - ie $a_r\sin\theta$.

• Ya you are right. Its a poorly worded problem of a conical pendulum. I confirmed. But instead of revolving T if you resolve mg along and perpendicular to the string. $mgcos\theta = T$ as there is no displacement in the direction of string I.e no slacking or extending of the string. 1.Now why is the tension I got different from your tension. 2. Then we get acceleration along the string is zero. 3. What happens to $mgsin\theta$? Its unbalanced and is it providing tangential acceleration to the mass? Feb 17, 2017 at 9:32
• Could you kindly solve it again by resolving mg instead of T? Feb 17, 2017 at 9:35
• $T=mg\cos\theta$ is incorrect because there is a non-zero component of centripetal acceleration along the string. There is no component of centripetal acceleration vertically, that is why I resolved vertically and horizontally instead of along and perpendicular to the string. The vertical forces on the ball are balanced, the horizontal forces are not. Feb 17, 2017 at 18:00

In polar coordinates: $P(r,\theta)$ As the string is inelastic: $$r=const. \quad \to \quad \dot{r}=\ddot{r}=0$$ $$\dot{\theta}=const.=36\frac{rad}{s} \quad \to \quad \ddot{\theta}=0$$

For curvilinear the acceleration towards the center $a_r$ is given by: $$a_r=\ddot{r}-r\cdot{\dot{\theta}}^2$$ $$\to a_r=-0.5m\cdot\left(36\frac{rad}{s}\right)^2=-648\frac{m}{s^2}$$