Tag: centre of mass

Questions Related to centre of mass

Two unequal masses are tied together with a cord with a compressed spring in between.
Which one is correct?

physics turning effects of forces stability and centre of mass center of mass centre of mass

  1. Both masses will have equal KE.

  2. Lighter block will have greater KE.

  3. Heavier block will have greater KE.

  4. None of above answers is correct.

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Correct Option: B
Explanation:
Lighter block will have greater kinetic energy to lighter block will have higher velocity mass so Heavier block, hence by equation $\dfrac{1}{2}mv^2,$ the lighter will have greater KE.
Hence, the answer is Lighter block will have greater KE.

A string is wrapped around a cylinder of mass $M$ and radius $R$. The string is pulled vertically upwards to prevent the centre of mass from falling as the cylinder unwinds the string, The work done on the cylinder for reaching an angular speed $\omega$ is:

physics turning effects of forces stability and centre of mass center of mass centre of mass

  1. $\cfrac { 2M{ R }^{ 2 }{ \omega }^{ 2 } }{ 3 } $

  2. $\cfrac { M{ R }^{ 2 }{ \omega }^{ 2 } }{ 3 } $

  3. $\cfrac { M{ R }^{ 2 }{ \omega }^{ 2 } }{ 2 } $

  4. $\cfrac { M{ R }^{ 2 }{ \omega }^{ 2 } }{ 4 } $

Show answer
Correct Option: D
Explanation:
Work done is the rotational KE acquired be cylinder,
$=\dfrac{1}{2} I\omega ^2$
$=\dfrac{1}{2}\dfrac{MR^2}{2}\omega ^2$
$=\dfrac{MR^2}{4}\omega ^2.$
Hence, the answer is $\dfrac{MR^2}{4}\omega ^2.$

A straight rod of length L has one of its ends at the origin and the other at $x=L$. If the mass per unit length of the rod is given by Ax where A is constant, where is its mass centre?

physics turning effects of forces stability and centre of mass center of mass centre of mass

  1. $L/3$

  2. $L/2$

  3. $2L/3$

  4. $3L/4$

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Correct Option: B
Explanation:
I assume you meant to say "a is a CONSTANT".

xc = coordinate of center of mass

M = total mass

$xc = ∫xdm / ∫dm = ∫xdm / M$

Given:$ m(x) = ax ⇒ dm/dx = a ⇒ dm = adx$

$∫xdm = ∫01 x(adx) = a/2$

$M = ∫dm = ∫(dm/dx)dx = ∫01 adx = a$

$xc = (a/2)/a = 1/2$

By the way, this is a mechanics problem (in statics), not a thermodynamics problem.

 

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is $\alpha R$ fromthe centre of the bigger disc. The value of $\alpha$ is

physics turning effects of forces stability and centre of mass center of mass centre of mass

  1. $\cfrac{1}{2}$

  2. $\cfrac{1}{6}$

  3. $\cfrac{1}{4}$

  4. $\cfrac{1}{3}$

Show answer
Correct Option: D