In a playground,there is a merry-go-round of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The initial angular momentum of the system is provided by the child running tangent to the merry-go-round. The merry-go-round is initially at rest

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) The initial angular momentum of the system is provided by the child running tangent to the merry-go-round. The merry-go-round is initially at rest.Initial angular momentum $L_i = mvr$,where $m = 30\ kg$,$v = 5\ m/s$,and $r = 4\ m$.$L_i = 30 	imes 5 	imes 4 = 600\ kg \cdot m^2/s$.After the child jumps on,the system rotates with angular velocity $\omega$. The total moment of inertia $I_{total} = I_{merry-go-round} + I_{child}$.$I_{merry-go-round} = Mk^2 = 120 	imes (3)^2 = 120 	imes 9 = 1080\ kg \cdot m^2$.$I_{child} = mr^2 = 30 	imes (4)^2 = 30 	imes 16 = 480\ kg \cdot m^2$.$I_{total} = 1080 + 480 = 1560\ kg \cdot m^2$.By conservation of angular momentum,$L_i = L_f = I_{total} \omega$.$600 = 1560 	imes \omega$.$\omega = 600 / 1560 = 60 / 156 = 10 / 26 = 5 / 13 \approx 0.3846\ rad/s$.Rounding to the nearest provided option,the value is $0.4\ rad/s$.

Similar Questions

$A$ cube of side '$a$' moves with a velocity '$v$' on a horizontal surface as shown in the figure. It hits an edge at point '$O$'. What will be the angular speed of the block after it hits point '$O$'?

$A$ force $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$ is acting at a point $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$. The value of $\alpha$ for which angular momentum about the origin is conserved is

The angular momentum of a system of particles is conserved:

Vedclass Test Series

Exam Paper Generator

Online Exam Module