Question

A copper block rests 43.4 cm from the center of a steel turntable. The coefficient of static friction between block and surface is 0.318. The turntable starts from rest and rotates with a constant angular acceleration of 0.259 rad/s 2 . The acceleration of gravity is 9.8 m/s 2 . After what interval will the block start to slip on the turntable?[ Hint: Ignore the tangential component of the acceleration.] Answer in units of s.

Answer 1

Answer:

10.3 s

Explanation:

We are given that

r=43.4 cm=$\frac{43.4}{100}=0.434 m$

1 m=100 cm

Coefficient of static friction=$\mu_s=0.318$

Angular acceleration=$\alpha=0.259rad/s^2$

Initial angular velocity =$\omega_1=0$

Acceleration of gravity=$g=9.8m/s^2$

Angular acceleration=$\alpha=\frac{\omega}{t}$

Using the formula

$0.259t=\omega$

Force acting on block after time t=$F=m\omega^2 r$

Friction force=$f=\mu mg$

$f=F$

$\mu mg=m\omega^2 r$

$\mu g=\omega^2 r$

Substitute the values

$0.318\times 9.8=(0.259t)^2 r$

$t^2=\frac{0.318\times 9.8}{(0.259)^2\times 0.434}$

$t=\sqrt{\frac{0.318\times 9.8}{(0.259)^2\times 0.434}}$

$t=10.3 s$

Hence, the block will start to slip on the turntable after 10.3 s.