Question

A coin has a weight of 100 grams and rotates on a horizontal disc with frequency of 1 hertz. The coin is at a distance of 10 cm from the axis of the disc. What is the frictional force that acts on the coin?
How much is the frictional force constant between the coin and the disc , when the coin is at 16 cm from the centre of rotation of the disc?

Answer 1

Answer on Question 63633, Physics, Mechanics, Relativity

Question:

A coin has a weight of 100 $g$ and rotates on a horizontal disc with frequency of 1 $Hz$. The coin is at a distance of 10 $cm$ from the axis of the disc. What is the frictional force that acts on the coin? How much is the frictional force constant between the coin and the disc, when the coin is at 16 $cm$ from the centre of rotation of the disc?

Solution:

a) Since the frictional force that acts on the coin provides the centripetal force, we get:

$$F_{fr} = F_{c},$$ $$F_{fr} = \frac{m v^{2}}{r},$$

here, $m$ is the mass of the coin, $v$ is the tangential velocity of the coin, $r$ is the distance from the axis of the disc to the coin.

Taking into account the relation between the distance from the axis, tangential velocity and the angular frequency of the rotation ($v = \omega r$) and substituting this relation into the previous equation we get:

$$F_{fr} = \frac{m v^{2}}{r} = \frac{m \omega^{2} r^{2}}{r} = m \omega^{2} r,$$

here, $\omega$ is angular frequency of the rotation of the coin.

There is a relation between the angular frequency and the ordinary frequency:

$$\omega = 2 \pi f.$$

Then, substituting it into the last formula we can calculate the frictional force that acts on the coin:

$$F_{fr} = m \omega^{2} r = m (2 \pi f)^{2} r = 0.1 \, kg \cdot 4 \cdot \pi^{2} \cdot \left(1 \, \frac{rev}{s}\right)^{2} \cdot 0.1 \, m = 0.395 \, N.$$

b) As in the previous case, we have:

$$F_{fr} = F_{c},$$ $$\mu m g = \frac {m v ^ {2}}{r},$$ $$\mu m g = m \omega^ {2} r,$$ $$\mu g = (2 \pi f) ^ {2} r.$$

From this formula we can find the frictional force constant between the coin and the disc:

$$\mu = \frac {(2 \pi f) ^ {2} r}{g} = \frac {4 \cdot \pi^ {2} \cdot \left(1 \frac {r e v}{s}\right) ^ {2} \cdot 0 . 1 6 m}{9 . 8 \frac {m}{s ^ {2}}} = 0. 6 5.$$

Answer:

a) $F_{fr} = 0.395N$

b) $\mu = 0.65$

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