# point masses m1 and m2 are placed at the opposite ends

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## Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum, is given by:

Click here👆to get an answer to your question ✍️ Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum, is given by:

Point masses mQuestion 1 and m 2

are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω

0

is minimum, is given by:

**A**x=

m 1 +m 2 m 2 L

**B**x=

m 1 +m 2 m 1 L

**C**x=

m 2 m 1 L

**D**x=

m 1 m 2 L Medium NEET Open in App Solution Verified by Toppr

Correct option is A)

MI of m1 about the axis: I 1 =m 1 x 2 MI of m 2 about the axis: I 2 =m 2 (L−x) 2 KE is rotational. Total KE is E= 2 1 I 1 ω 0 2 + 2 1 I 2 ω 0 2 = 2 1 ω 0 2 (m 1 x 2 +m 2 (L−x) 2 )

Work done is change in KE.

To minimize E, differentiate wrt x and equate to zero.

m 1 x−m 2 (L−x)=0 ⇒x= m 1 +m 2 m 2 L

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at

m 1 +m 2 m 2 L Video Explanation

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## Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L,and negligeable mass. The rod is set rotating about an axis perpendicular to it. The position of point P on this rod through which axis should pass so that the work required to set the rod rotating with angular velocity w0(omega) is minimum, is given by

Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L,and negligeable mass. The rod is set rotating about an axis perpendicular to it. The position of point P on this rod through which axis should pass so that the work required to set the rod rotating with angular velocity w0(omega) is minimum, is given by

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Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L,and negligeable mass. The rod is set rotating about an axis perpendicular to it. The position of point P on this rod through which axis should pass so that the work required to set the rod rotating with angular velocity w0(omega) is minimum, is given by

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Open in App Solution

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SIMILAR QUESTIONS

**Q.**Point masses

m 1 and m 2

are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity

ω 0

is minimum, is given by:

**Q.**Point masses

m 1 and m 2

are placed at the opposite ends of a rigid rod of length

L

of negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point 'P' on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity

ω 0

is minimum is given by

**Q.**Point masses

m 1 and m 2

are placed at the opposite ends of a rigid rod of length

L

of negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point 'P' on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity

ω 0

is minimum is given by

**Q.**Two point masses of

0.3 k g and 0.7 k g

are fixed at the ends of a rod of length

1.4 m

and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of

**Q.**A thin rod of negligible mass and length 1.4 m has two point masses of 0.3. kg and 0.7 kg are fixed at its ends.

The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. In order that the work required for rotation of the rod be minimum, the point on the rod through which the axis should pass, is located from the 0.3 kg mass at a distance of

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## Point masses m(1) and m(2) are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicualr to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity omega(0) is minimum, is given by :

Let the axis of rotation be at a distance x from m(1) Fig K.E. of translation = 0 K.E. of rotation = (1)/(2) (I(1) + I(2))omega(0)^(2) According to work energy principle, Work done = Increase in energy W = (1)/(2) (I(1) + I(2)) omega(0)^(2) = (1)/(2) [m(1)x^(2) + m(2) (L - x)^(2)]omega(0)^(2) ..(i) For W to be minimum,(dW)/(dx) = 0 From (i), (dW)/(dx) = (1)/(2)m(1)2x + (1)/(2)m(2) xx 2(L - x) = 0 or m(1) x - m(2) (L - x) = 0 or (m(1) + m(2)) x = m(2)L x = m(2)L//(m(1) + m(2)) This is the position of centre of mass of the ord from m(1). Hence the required axis should pass through centre of mass of the rod and perpendicular to the length of the rod as shown in Fig.

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Systems Of Particles And Rotational Motion

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Point masses m_(1) and m_(2) a...

Point masses m 1 and m 2 m1andm2

are placed at the opposite ends of a rigid rod of length

L L

, and negligible mass. The rod is to be set rotating about an axis perpendicualr to it. The position of point

P P

on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity

ω 0 ω0

is minimum, is given by :

Updated On: 27-06-2022

00 : 010

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Text Solution Open Answer in App Solution

Let the axis of rotation be at a distance x from

m 1 m1 Fig K.E. K.E. of translation =0 =0 K.E. K.E. of rotation = 1 2 ( I 1 + I 2 ) ω 2 0 =12(I1+I2)ω02

According to work energy principle,

Work done = Increase in energy

W= 1 2 ( I 1 + I 2 ) ω 2 0 = 1 2 [ m 1 x 2 + m 2 (L−x) 2 ]ω (0) 2

W=12(I1+I2)ω02=12[m1x2+m2(L-x)2]ω(0)2

..(i) For W W to be minimum, dW dx =0 dWdx=0 From (i), dW dx = 1 2 m 1 2x+ 1 2 m 2 ×2(L−x)=0or m 1 x− m 2 (L−x)=0or( m 1 + m 2 )x= m 2 L

dWdx=12m12x+12m2×2(L-x)=0orm1x-m2(L-x)=0or(m1+m2)x=m2L

x= m 2 L/( m 1 + m 2 ) x=m2L/(m1+m2)

This is the position of centre of mass of the ord from

m 1 m1

. Hence the required axis should pass through centre of mass of the rod and perpendicular to the length of the rod as shown in Fig.

Answer

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Guys, does anyone know the answer?