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Rotational Motion question

2007 · Shift 2 · Q18
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  5. /2007 · Shift 2 · Q18

Rotational Motion question

2007 · Shift 2 · Q18

JEE AdvancedPhysicsRotational MotionMCQ+3 / −1
STATEMENT 1 If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant. Because STATEMENT 2 The linear momentum of an isolated system remains constant.
  1. A
    Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
  2. B
    Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
  3. C
    Statement-1 is True, Statement-2 is False.
  4. D
    Statement-1 is False, Statement-2 is True.
View written solutionFree

Correct answer: D

Step-by-step analysis:

  1. Analyze Statement 1:

    • The statement is: "If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant."
    • The velocity of the center of mass, v⃗CM\vec{v}_{CM}vCM​, of a body remains constant if and only if its acceleration, a⃗CM\vec{a}_{CM}aCM​, is zero.
    • According to Newton's second law for translational motion of a system, the net external force F⃗ext\vec{F}_{ext}Fext​ is related to the acceleration of the center of mass by F⃗ext=Ma⃗CM\vec{F}_{ext} = M \vec{a}_{CM}Fext​=MaCM​, where MMM is the total mass.
    • Thus, for v⃗CM\vec{v}_{CM}vCM​ to be constant, the net external force F⃗ext\vec{F}_{ext}Fext​ must be zero.
    • Statement 1 claims that if the net external torque about the center of mass ( auCM\ au_{CM} auCM​) is zero, then v⃗CM\vec{v}_{CM}vCM​ is constant. This implies that if  auCM=0\ au_{CM} = 0 auCM​=0, then F⃗ext=0\vec{F}_{ext} = 0Fext​=0.
    • Let's consider a counterexample. Suppose a non-zero net external force F⃗ext\vec{F}_{ext}Fext​ acts on the body, and its point of application is the center of mass itself.
    • The torque about the center of mass is given by τ⃗CM=r⃗CM×F⃗ext\vec{\tau}_{CM} = \vec{r}_{CM} \times \vec{F}_{ext}τCM​=rCM​×Fext​, where r⃗CM\vec{r}_{CM}rCM​ is the position vector of the point of application of the force relative to the center of mass. In this case, r⃗CM=0\vec{r}_{CM} = 0rCM​=0.
    • Therefore, the torque about the center of mass is τ⃗CM=0×F⃗ext=0\vec{\tau}_{CM} = 0 \times \vec{F}_{ext} = 0τCM​=0×Fext​=0. So the condition in the statement is met.
    • However, since F⃗ext\vec{F}_{ext}Fext​ is not zero, the center of mass will accelerate: a⃗CM=F⃗extM≠0\vec{a}_{CM} = \frac{\vec{F}_{ext}}{M} \neq 0aCM​=MFext​​=0. This means the velocity of the center of mass, v⃗CM\vec{v}_{CM}vCM​, will change.
    • Since we found a situation where the premise ( auCM=0\ au_{CM} = 0 auCM​=0) is true but the conclusion ("vCM"v_{CM}"vCM​ is constant") is false, Statement 1 is False.
  2. Analyze Statement 2:

    • The statement is: "The linear momentum of an isolated system remains constant."
    • An 'isolated system' is defined in physics as a system on which the net external force is zero (F¨ext=0\"{F}_{ext} = 0F¨ext​=0).
    • Newton's second law states that the net external force on a system is equal to the rate of change of its total linear momentum (p⃗\vec{p}p​): F⃗ext=dp⃗dt\vec{F}_{ext} = \frac{d\vec{p}}{dt}Fext​=dtdp​​.
    • If the system is isolated, F⃗ext=0\vec{F}_{ext} = 0Fext​=0. Substituting this into the equation gives dp⃗dt=0\frac{d\vec{p}}{dt} = 0dtdp​​=0.
    • This implies that the total linear momentum p⃗\vec{p}p​ of the system must be a constant vector. This is the well-known principle of conservation of linear momentum.
    • Therefore, Statement 2 is True.
  3. Conclusion:

    • We have determined that Statement 1 is False and Statement 2 is True.
    • Looking at the options, this corresponds to option D.

Final Answer Selection:

  • Statement-1 is False.
  • Statement-2 is True.
  • This matches option D.
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