Practice free Rotational Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A wheel rotates with a constant acceleration of $ 2.0 rad /sec ^ 2 $ If the wheel start from rest. The number of revolution it makes in the first ten seconds will be approximately.
$ \theta = wot + { 1 \over 2} \alpha t^2 \Rightarrow \theta = 100 rad $
Two discs of the same material and thickness have radii 0.2 m and 0.6 m their moment of inertia about their axes will be in the ratio
$ I = { 1 \over 2 } MR^2 = { 1 \over 2} ( \pi R^2 t \times \rho ) R^2 $ $ A t \times \rho are same $ $ I \alpha R^4 \therefore { I_1 \over I_2 } = \left( { R_1 \over R_2 } \right) ^4 $
A wheel of mass 10 kg has a moment of inertia of$ 160 kg m^2 $ about its own axis. The radius of gyration will be ………... m.
$ I = MK^2 = 160 $ $ \therefore K^2 = {160 \over m} = { 160 \over 10 } = 16 $ $ \therefore K = 4 $
One circular rig and one circular disc both are having the same mass and radius. The ratio of their moment of inertia about the axes passing through their centres and perpendicular to their planes, will be……
$ { I ring \over I disc } = { MR^2 \over { 1 \over 2 } MR^2 } = { 2 \over 1 } $ $ \therefore 2 :1 $
A ring of mass M and radius r is melted and then molded in to a sphere then the moment of inertia of the sphere will be…..
$ I_{ring} = MR _1^ 2$ As Volume and Mass remain same $ I_{solid } = { 2 \over 5} MR_2^2 $ $ R_2 \lt \lt R_1 $
A circular disc of radius R and thickness R/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is …
Volume of disc = $ V_1 = \pi R_1 ^2 . t = { \pi R_1^3 \over 6 } $ $ \therefore { R_1^3 \over 6} = { 4 \over 3} R_2 ^3 $ Volume of Sphere = $ = { 4\over 3} \pi R_2^3 $ $ R_1^3 = 8 R_2^3 $ $ \therefore R_1 = 2R_2 $ $ I_1 = M.I of disc = I = { 1 \over 2} MR_1^2 $ $ I_2 = M.I of sphere = { 2 \over 5} MR_2^2 = { mR_1^2 \over 5 \times 2 } = { I \over 5} $
Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio 1:3. The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio.
$M.I of disc = {1 \over 2} MR^2 = { 1 \over 2 } M \left( { M \over \pi \rho } \right) = { 1 \over 2} { M^2 \over \pi t \rho } $ As their mass & thickness are some $ I \alpha { 1 \over \rho } $
Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle of Q with AB. The moment of inertia of the plate about the axis CD is then equal to….
Let $I_Z $ be the M.I of square plote about the axis passing through the centre and perpendicular to the plane of square, hence according to Perfendicular axis theorm. $ I_Z = I _{AB } + I_{AB} Also I_Z = I_{CD} + I_{C'D'} $ As axis are symmetric $ I_{AB} = I_{A'B'} = { I_z \over 2 } $ And $I_{CD} = I_{C'D'} = { I_Z \over 2 } $ So we can say that $I_{AB} = I_{A'B'} = I_{CD} = I_{C'D'} =I $
A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of the disc
Let the radius of complete disc is a & that of small disc is b After small disc is cut from complete disc let the C.M. shift to $O_2$ at distance $x_2$ flem original centre O. The Position of new C.M. is givenly let 6 is mass percunitarea. $ X_ {cm} = { -6 \pi b^2 \times 1 \over 6 \pi a^2 - 6 \pi b^2 } $
A straight rod of length L has one of its ends at the origin and the other end at x=L If the mass per unit length of rod is given by Ax where A is constant where is its center of mass
Let the mass of an element of length dx of the rod located at a distance X away from left and is $ { M \over L } dx $ , the x cordinate of the C.M. is given by. $ Total mass of rod = \int_0^1 Ax .dx = {AL^2 \over 2 } $ $ x_{cm} = { 1 \over M } \int xdm = { 1 \over \left( { AL^2 \over 2 } \right)} \int _0^1 x ( Ax dx ) $
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