Question

A small particle of mass 0.36 g rest on a horizontal turntable at a distance 25 cm from the axis of spindle. The turntable is accelerated at a rate of a = 1/3 rad/s^–2. The frictional force that the table exerts on the particle 2 s after the startup is-

• A. 40 mN
• B. 30 mN
• C. 50 mN
• D. 60 mN

Answer 1

Answer: (C)

50 mN

Answer 2

f : friction force ; f : $\sqrt{f_{c}^{2} + f_{t}^{2}}$

After 2 sec. :- w = at = $\frac{2}{3}$rad/sec

So f_t = ma_t = m a R = 36×10^–5 ×$\frac{1}{3}$×$\frac{1}{4}$N

F_c = mw²R = 36 × 10^–5 ×$\frac{4}{9}$×$\frac{1}{4}$N