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System of Particles and Rotational Motion Test - 18

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System of Particles and Rotational Motion Test - 18
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  • Question 1
    1 / -0
    Two rings of radius $$R$$ and $$nR$$ made of same material have the ratio of moment of inertia about an axis passing through centre in $$1 : 8$$. The value of $$n$$ is
    Solution
    Ratio of moment of inertia of the rings
    $$\dfrac {I_{1}}{I_{2}} = \left (\dfrac {M_{1}}{M_{2}}\right )\left (\dfrac {R_{1}}{R_{2}}\right )^{2} = \left (\dfrac {\lambda L_{1}}{\lambda L_{2}}\right )^{2} \left (\dfrac {R_{1}}{R_{2}}\right )^{2}$$
    $$= \left (\dfrac {2\pi R}{2\pi nR}\right ) \left (\dfrac {R}{nR}\right )^{2}$$
    $$[\lambda =$$ linear density of wire $$= constant]$$
    $$\Rightarrow \dfrac {L_{1}}{L_{2}} + \dfrac {1}{n_{3}} + \dfrac {1}{8}$$ (given)
    $$\therefore n^{3} = 8$$
    $$\Rightarrow n = 2$$.

  • Question 2
    1 / -0
    A wheel of mass $$10  kg$$ has a moment of inertia of $$160  kg m^2$$ about its own axis. The radius of gyration is :
    Solution
    Given moment of inertia of the wheel about its axis is $$ 160 kg m^2$$

    $$I = MK^2 $$ (K is the radius of gyration)

    $$160 = 10 \times K^2 $$

    $$K = 4m $$

    Radius of gyration of the wheel about its own axis is 4 m.
  • Question 3
    1 / -0
    Consider the following two statements A and B and identify the correct choice
    A) The torques produced by two forces of couple are opposite to each other
    B) The direction of torque is always perpendicular to plane of rotation of body
    Solution
    The direction of torque is always perpendicular to plane of rotation of body as a cross product is in perpendicular plane to $$r$$ and $$F$$ vectors and the torques produced by two forces of couple are in same direction to each other.

  • Question 4
    1 / -0
    A particle is moving along a fixed circular orbit with uniform speed. Then the correct statement among the following is:
    Solution
    Let fixed orbit have radius 'R' and uniform velocity of the partial is V.
    According to Question:
    Angular momentum $$L=M (\vec V\times \vec R)$$ .......(1) 
    Direction $$=$$ Perpendicular to the plane of motion ...(2)
    $$(\because$$ V, R is constant so $$L$$ will be constant as well$$)$$.
    from equation 1,2. L will be constant in magnitude and direction.
    Option C

  • Question 5
    1 / -0
     A body of mass 'M' collides against a wall with a velocity 'v' and retraces its path with the same velocity, the change in momentum is
    Solution
    $$\Delta P = Final\:\:mom - Initial\:\:mom$$

    $$= Mv - ( - Mv)$$

    $$= 2 Mv$$
  • Question 6
    1 / -0
    A uniform thin rod of mass $$M$$, length $$L$$ is vertically straight on a horizontal floor. If the rod is falling to a side without slipping at its bottom, the moment of inertia is:
    Solution

    MI of a rod of length L and mass M about an axis passing through one end is,
     $$I= \dfrac{ML^2}{3}$$
    (since body is not slipping so inertia will be considered from one end of the rod)

  • Question 7
    1 / -0
    Which of the following pairs of physical quantities are not analogous to each other in case of translatory motion and rotational motion, 
    Solution
    $$Couple$$ and $$Torque$$ are actually in the same context because both are produced due to the application of force. Moreover both are the faces of same coin.Hence they are not analogous to each other.
  • Question 8
    1 / -0
    Consider the following two statements A and B and identify the correct choice.
    A) The moment of inertia of a rigid body is independent of its angular velocity.
    B) The radius of gyration of a rotating metallic disc is dependent on its temperature.
    Solution
    (A) Moment of inertia of the rigid body only dependent on the mass of the body and position of the axis of rotation. It is independent of the angular velocity.
    (B) $$I =mk^{2}$$

    $$k=\sqrt{\dfrac{T}{m}}   (k=$$radius of gyration)

    $$I_{dist}=\dfrac{mR^{2}}{2}$$

    $$k=\dfrac{R}{\sqrt{2}}$$

    and Radius of the metallic disc is dependent on the temperature, because

    $$A=A_{0}(1+B\Delta T)$$   (Equation of thermal expansion)

    $$\pi r^{2}=\pi R_{0}^{2}(1+B\Delta T)$$

    $$r=R_{0}\sqrt{1+B\Delta T}$$
  • Question 9
    1 / -0
    The centre of a mass of a rigid body lies
    Solution
    Centre of mass of rigid bodies may lie either inside or outside the body.
    For example:  COM of solid sphere lies inside it whereas COM of uniform circular ring lies at its geometric centre where there is actually no matter.
  • Question 10
    1 / -0
    Can the centre of gravity be situated outside the material of the body ?
    Solution
    Yes, it can. For example, in case of a ring, it is situated at the centre of that circle. But the material is only along the circumference. Hence centre of gravity is situated outside the material of the body.
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