Question

A small coin of mass 80g is placed on the horizontal surface of rotating disc. The disc starts from rest and is given a constant angular acceleration = 2 rad/s^2. The coefficient of statis friction between the coin and the disc is u=3/4.The magnitude of resultant force is

Answer 1

Answer: The magnitude of resultant force is $F = 0.44 N$.

Explanation:

Given that,

Mass of coin m = 80 g

Angular acceleration $\alpha = 2 rad/s^{2}$

Coefficient of statics friction $\mu = \dfrac{3}{4}$

The frictional force is

$F_{f} = \mu mg$

$F_{f} = \dfrac{3}{4}\times0.08 kg \times 9.8 m/s^{2}$

$F_{f} = 0.6 N$

Now, the resultant force is

$F+ F_{f} = m\alpha$

$F = (0.08\times 2)N - 0.6 N$

$F = -0.44 N$

Negative sign indicates the resultant force acting in the opposite direction of the frictional force.

Hence, the magnitude of resultant force is $F = 0.44 N$.

Answer 2

"Mass of coin, m = 80 g

Angular acceleration, $\alpha \quad =\quad 2 \frac { rad }{ { s }^{ 2 } }$

Coefficient of statics friction, $\mu \quad =\quad \frac { 3 }{ 4 }$

The frictional force is,

${ F }_{ f }\quad =\quad \mu mg$

${ F }_{ f }\quad =\quad \frac { 3 }{ 4 } \quad \times \quad 0.08 Kg\quad \times \quad 9.8 \frac { m }{ { s }^{ 2 } }$

${ F }_{ f }\quad =\quad 0.6\quad N$

Now, the resultant force is,

$F\quad +\quad { F }_{ f }\quad =\quad m\alpha$

$F\quad =\quad \left( 0.08\quad \times \quad 2 \right) N\quad -\quad 0.6\quad N$

$F\quad =\quad -0.44\quad N$

Negative sign indicates the resultant force acting in the opposite direction of the frictional force.

Hence, the magnitude of resultant force, $F\quad =\quad -0.44\quad N$"