Question

On a horizontal disc, a block of mass $1kg$ is placed at a distance $3cm$ from the centre of the disc. The contact surface between the disc and the block is rough having ${\mu }_s=1/2$. Now the disc is rotated with a constant angular velocity of $10rad/s$ about its own axis. The contact force applied by the disc on the block is

• A. $\sqrt{109}N$
• B. $\sqrt{125}N$
• C. $3N$
• D. $10N$

Answer 1

Answer: (A)

$\sqrt{109}N$

Centripetal force $F=m{\omega }^{2}r=1\times {10}^2\times 3\times {10}^{-3}=3N$ (towards center)

Normal force $N^{\prime }=mg=1\times 10=10N$ (upwards)

Since $F$ and $N^{\prime }$ are pendicular to each other.

Total contact force $R=\sqrt{F^2+(N^{\prime }{)}^2}=\sqrt{{10}^2+3^2}=\sqrt{109}N$