Tag: center of mass

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Questions Related to center of mass

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

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

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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:

  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 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac{MR^2}{4}\omega ^2.$

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

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?

  1. $L/3$

  2. $L/2$

  3. $2L/3$

  4. $3L/4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

 

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

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

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

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

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

  4. $\cfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the negative mass method, the center of mass of the remaining part is found by subtracting the mass and moment of the smaller disc from the larger one. The calculation leads to alpha = 1/3.