Question

A child with mass 40 kg sits on the edge of a merry-go-round at a distance of 3.0 m from its axis of rotation. The merry-go-round accelerates from rest up to 0.4 rev/s in 10 s. If the coefficient of static friction between the child and the surface of the merry-go-round is 0.6, does the child fall off before 5 s?

Answer 1

Answer:No

Explanation:

Given

mass $m=40 kg$

$N=0.4 rev/s$

$\omega =0.8\pi rad/s$

time $t=10 s$

using

$\omega =\omega _0+\alpha \cdot t$

$0.8\pi =\alpha 10$

$\alpha =0.08\pi rad/s^2$

at t=5 s

$\omega _2=0+\alpha 5$

$\omega _2=0.4\pi rad/s$

tangential acceleration $a_t=\alpha r$

$a_t=0.08\pi \times 3=0.08\pi=0.754 m/s^2$

centripetal acceleration $a_c=(\omega _2)^2r$

$a_c=(0.4\pi )^{2}\times 3=4.73 m/s^2$

$a_{net}=\sqrt{(a_c)^2+(a_t)^2}$

$a_{net}=\sqrt{(4.73)^2+(0.754)^2}$

$a_{net}=\sqrt{22.941}$

$a_{net}=4.789 m/s^2$

maximum frictional force$=\mu mg$

$F_r=0.6\times 40\times 10=235.2 N$

maximum Force due $a_{net}=m\times a_{net}$

$=40\times 4.789=191.58 N$

$F_s> ma$

Therefore child does not fall