# Circular motion, kinetic friction and tension

Hi: I've read this post: Uniform Circular Motion w/ Tension and Friction and helped me a lot.

I have a similar problem:

At one end of the rope ($R = 1 m$) is tied a mass ($m = 3 kg$) and the other end is attached to a spherical joint located in the center of a entirely horizontal turntable. The mass is on the turntable and both remain at rest ($V = 0 m/s$ , $\omega = 0 rad/s$). When the turntable begins to rotate, the mass remains without relative velocity on the turntable, rotating at an unknown speed until its acceleration increases and the mass begins to have an angular velocity different from that of the turntable. The kinetic coefficient between the mass and the turntable is 0.1 and the maximum tension of the string is $100 N$. Calculate the time required by the mass to reach enough speed to break the rope.

I analyzed the problem and this is what I think:

The turntable moves and the mass rotate together (there is static friction) at an unknown speed. Then, the mass has a relative motion due to the acceleration and the kinetic friction (between the turntable and the mass). Finally the mass reach a speed that make the rope breaks and the mass leaves the turntable.

I think that when the tension is maximum, I can use this equation to describe the tension:

$$T = m\frac{V^2}{R}$$

Based in the previous formula, I can use it to calculate the speed of the mass at that instant of the following equation:

$$V = \sqrt{\frac{TR}{m}}$$

And to find the time in which the mass exits the turntable with the equation:

$$t = \frac{2\pi R}{V}$$ (I'm not sure)

I know the kinetic friction acts in opposite direction of the tangential acceleration and perpendicular to the tension and the centripetal acceleration.

The tangential acceleration and the angular acceleration ($\alpha$) are unknown and any of them can help to solve the problem: how long the mass rotate until the rope breaks?

• Can you clarify the setup? Are you starting by assuming the mass is initially stationary with respect to the turntable, i.e. rotating with it, and held at a fixed distance from the centre by the rope? Then the rope breaks? – John Rennie Feb 17 '17 at 7:13
• @JohnRennie I corrected the question – Diana Arosemena Feb 17 '17 at 8:26
• Ah, so you mean the mass starts stationary and sliding on the turntable. Then the (kinetic) friction between the mass and the turntable accelerates the mass until it is moving so fast that the string breaks. Is that correct? – John Rennie Feb 17 '17 at 8:33
• @JohnRennie yes, that's correct. – Diana Arosemena Feb 17 '17 at 16:19
• Your "conceptual question" ("am I using the concepts correctly") is really the same as asking "Is my solution correct?" Your final question ("does kinetic friction affect speed") has a trivial answer : yes, if there is relative motion then kinetic friction tends to reduce relative speed. – sammy gerbil Feb 17 '17 at 18:22

In this problem there are 2 forces providing centripetal (radial) acceleration : the tension $T$ in the rope, and the radial component $f_r$ of the static friction force between the block and the turntable. There is only one force causing tangential acceleration : the tangential component $f_t$ of the static friction force.

Note that the static friction force does not have to be radial or tangential. The resultant $f=\sqrt{f_r^2+f_t^2}$ must satisfy the constraint $f \le \mu mg$ where $\mu$ is the coefficient of limiting static friction.

Your problem statement mentions only kinetic friction not static, but the question Can the coefficient of static friction be less than that of kinetic friction? shows that static friction must be at least equal to kinetic friction.

The tangential acceleration of the block is $r\alpha$ so $f_t=mr\alpha$.

When the turntable and block have reached angular speed $\omega$ then the centripetal force required to keep it in circular motion is $mr\omega^2$. When the rope is on the point of breaking then
$T_{max}+f_r=mr\omega^2$.

If $f_t, f_r$ are comparable then
$f^2=f_r^2+f_t^2=(mr\omega^2-T_{max})^2+(mr\alpha)^2 \le (\mu mg)^2$.
Solving this equation gives you the value of $\omega$ at which the rope will break. The time $t$ after acceleration started is given by $\omega=\alpha t$, assuming that the acceleration is constant.

I think you are probably expected to simplify the problem by assuming that $\alpha$ is small so that $f_t \ll f_r$. Then you need to solve
$mr\omega^2 \le T_{max}+\mu mg$.