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151.

A right circular cylinder of radius r and mass m is suspended by a cord that is wound around its surface. It is allowed to fall, prove that at its center of gravity will follow a vertical rectilinear path and find the accleration along this path. Determine also the tensile force in the cord.

Answer» Correct Answer - `(2)/(3)g;(1)/(3)mg`
Discussion
152.

Figure shows a counterweight of mass `m` suspended by a cord wound around a spool of radius `r`. forming part of a turntable supporting the object. The turntable can rotate without friction. When the counterweight is released from rest, it descends through a distance `h`, acquiring a speed `v`. The moment of inertia `I` of the rotating apparatus is. .A. `mr^2((2gh)/(v^2) +1)`B. `mr^2((g h)/(v^2) +1)`C. `mr^2((2 g h)/(v^2)-1)`D. `mr^2((g h)/(v^2)+1)`

Answer» Correct Answer - C
( c) Each point on the cord moves at a linear speed of `v = omega r` where `r` is the radius of the spool.The energy conservation equation for the counterweight-turntable-Earth system is :
`(K_1 + K_2 + U_g)_i + |W_(other) = (K_1 + K_2 + U_g) _f`
Specializing, we have
`0 + 0 + mgh + 0 = (1)/(2)mv^2 + (1)/(2) I omega^2 + 0`
`mgh = (1)/(2) = (1)/(2) mv^2 + (1)/(2) I (v^2)/(r^2)`
`2mgh - mv^2 = I (v^2)/(r^2)` and finally, `I = mr^2 ((2gh)/(v^2) - 1)`.
Discussion
153.

A `60.0-kg` woman stands at the western rim of a horizontal turntable having a moment of inertia of `50 kg.m^2` and radius of `2.0 m`. The turntable is initially at rest and free to rotate abot a frictionless, vertical axle through its centre. The woman then starts walking around the rim clockwise` ("as viewed from above the system")` at constant speed of `1.50 m//s` relative to the Earth. The final angular velocity of the woman and the turntable systems.A. `0.36 rad//s ("counterclockwise")`B. `1.8 rad//s ("counterclockwise")`C. `3.6 rad//s ("clockwise")`D. `0.36 rad//s("clockwise")`

Answer» Correct Answer - A
(a) From conservation of angularmomentum for the system of the woman and the turntable, we have `L_f = L_i = 0`,
So `L_f = I_(woman) omega_("woman") + I_("table") omega_("table") = 0`
and `omega_("table") = (-(I_(woman))/(I_("table"))) omega_(woman)`
=`((m_(woman) r^2)/(I_("table")))(v_(woman)/(r))`
=`(m_(woman) rv_(woman))/(I_("table"))`
`omega_("table") = -(60.0 kg(2.00 m)(1.50 m//s))/(500 kg .m^2)`
=` -0.360 rad//s`
or `omega_("table") = 0.360 rad//s ("counterclockwise")`.
Discussion
154.

Two uniform spheres of mass `M` have radii `R` and `2 R`. Each sphere is rotating about a fixed axis through a diameter, the rotational kinetic energies of the sphere are identical. What is the ratio of the magnitude of the angular momentum of these spheres gt That is, `(L_(2 R))/(L_( R)) =`A. 4B. `2 sqrt(2)`C. 2D. `sqrt(2)`

Answer» Correct Answer - C
`KE = (L^(2))/(2I)`
`L = sqrt(2I(K.E)) rArr (L_(2R))/(L_(R)) = sqrt((I_(2R))/(I_(R))) = sqrt(((2)/(5) xx m(2R)^(2))/((2)/(5) mR^(2)))`
`(L_(2R))/(L_(R)) = (2)/(1)`.
Discussion
155.

There object, `A` : (a solid sphere), `B` : (a thin circular disk) and `C` : (a circular ring), each have the same mass `M` and radius `R`. They all spin with the same angular speed `omega` about their own symmetry axes. The amount of work `(W)` required ot bring them to rest, would satisfy the relationA. `W_A gt W_C gt W_B`B. `W_C gt W_B gt W_A`C. `W_B gt W_A gt W_C`D. `W_A gt W_B gt W_C`

Answer» Correct Answer - B
(b) Using work energy theorem
`W = Delta K = (1)/(2) I omega^2`
`W prop I`
`W_A : W_B : W_C = (2)/(5) MR^2 : (1)/(2) MR^2 : MR^2`
`W_A : W_B : W_C = (2)/(5) : (1)/(2) : 1`
Hence `W_C gt W_B gt W_A`.
Discussion