Question

2 blocks of masses m1=10kg and m2=5kg connected to each other by a massless in extensible string of length 0.3m and placed along the diameter of a turn table coefficient of friction=0.5 between surface of table and m1 and there is no friction between table and m2.the table is rotating with 10rad/s about the vertical axis.the mass m1 is at a distance of 0.124m from centre.1)if the masses are at rest,calculate the frictional force on m1.2)what should be the angular speed of table if the masses just start to slip?

Answer 1

Answer on Question#57690 - Physics - Mechanics - Relativity

2 blocks of masses $m_{1} = 10\mathrm{kg}$ and $m_{2} = 5\mathrm{kg}$ connected to each other by a massless in extensible string of length $L = 0.3\mathrm{m}$ and placed along the diameter of a turn table coefficient of friction $\mu_{1} = 0.5$ between surface of table and $m_{1}$ and there is no friction between table and $m_{2}$. The table is rotating with $\omega = 10\frac{\mathrm{rad}}{\mathrm{s}}$ about the vertical axis. The mass $m_{1}$ is at a distance of $l_{1} = 0.124\mathrm{m}$ from centre.

1) If the masses are at rest, calculate the frictional force on $m_{1}$.

2) What should be the angular speed of table if the masses just start to slip?

Solution:

1) The centrifugal force acting on $m_{1}$ is

$$F _ {1} = m _ {1} \omega^ {2} l$$

The centrifugal force acting on $m_{2}$ is

$$F _ {2} = m _ {2} \omega^ {2} (L - l)$$

The frictional force on $m_{1}$ is equal to the difference of the above forces:

$$\begin{array}{l} F _ {f} = \left| F _ {2} - F _ {1} \right| = \omega^ {2} \left| m _ {2} (L - l) - m _ {1} l \right| = \\ = \left(1 0 \frac {\mathrm {r a d}}{\mathrm {s}}\right) ^ {2} | 5 \mathrm {k g} \cdot (0. 3 \mathrm {m} - 0. 1 2 4 \mathrm {m}) - 1 0 \mathrm {k g} \cdot 0. 1 2 4 \mathrm {m} | = 3 6 \mathrm {N} \\ \end{array}$$

2) If masses start to slip, the force $F_{f}$ must be equal to the force of kinetic friction $F_{f}^{k} = m_{1}g\mu_{1}$ acting on the mass $m_{1}$:

$$\begin{array}{l} F _ {f} ^ {k} = F _ {f} \\ m _ {1} g \mu_ {1} = \omega^ {2} \left| m _ {2} (L - l) - m _ {1} l \right| \\ \end{array}$$

Thus,

$$\omega = \sqrt {\frac {m _ {1} g \mu_ {1}}{| m _ {2} (L - l) - m _ {1} l |}} = \sqrt {\frac {1 0 \mathrm {k g} \cdot 9 . 8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot 0 . 5}{| 5 \mathrm {k g} \cdot (0 . 3 \mathrm {m} - 0 . 1 2 4 \mathrm {m}) - 1 0 \mathrm {k g} \cdot 0 . 1 2 4 \mathrm {m} |}} = 1 1. 7 \frac {\mathrm {r a d}}{\mathrm {s}}$$

Answer:

1) 36N

2) $11.7\frac{\mathrm{rad}}{\mathrm{s}}$

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