Question
A child sits stationary at one end of a long trolley moving uniformly with a speed $V$ on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Step 1
The center of mass of a system of particles is the point that moves as though all of the system's mass were concentrated there and all external forces were applied there. Show more…
Show all steps
Your feedback will help us improve your experience
Sanu Kumar and 58 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A child is standing at one end of a long trolley moving with a speed $v$ on a smooth horizontal floor. If the child starts running towards the other end of the trolley with a speed $u$, the centre of mass of the system (trolley + child) will move with a speed (a) zero (b) $(v+u)$ (c) $(v-u)$ (d) $v$
A child is standing at one end of a long trolley moving with a speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u, the centre of mass of the system (trolley + child) will move with a speed. A. Zero B. (v + u) C. v D. (v - u)
Physics. The speed of a child riding a merry-go-round at a carnival is given by the equation $v=\sqrt{12 r},$ where $v$ is the speed in feet per second and $r$ is the distance in feet from the center of the merry-go-round to the rider. If a child is moving at $15 \mathrm{ft} / \mathrm{s}$ how far is the child from the center of the merry-go-round?
Radical Expressions
Solving Equations Containing Radical Expressions
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD