In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is sufficiently rough to ensure rolling, what is the kinetic energy of the body now in the given time interval of `2s`?A. `18.75J`B. `25.67J`C. `16.67J`D. none of these

In the given figure `F=10N, R=1m`, mass of the body is `2kg` and momen

1.	In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is sufficiently rough to ensure rolling, what is the kinetic energy of the body now in the given time interval of `2s`?A. `18.75J`B. `25.67J`C. `16.67J`D. none of these
Answer» Correct Answer - C When frictional force is absent, the rigid body does not translate `SigmavecF_("external")=0` therefore `veca_(CM)=0` Taking torque about `CM` `Fxx2R-FxxR=I_(CM)alpha` `10xx1=4alpha` `alpha=5/2rad//s^(2)` `omega=alphat=5rad//s` `KE_("total")=1/2I_(CM)omega^(2)=1/2xx4xx5xx5=50J` Taking torque about `IC` (bottom most point) `Fxx4R-Fxx3R=[I_(CM)+M(2R)^(2)]alpha` `=10xx1=[4+2xx(2xx1)^(2)]alpha` `alpha=10/2rad//s^(2)` `omega=alphat=10/12xx2=5/2rad//s` `KE_("total")=1/2[I_CM+M(2R)^(2)]omega^(2)` `=1/2[4+2xx(2xx2)^(2)]xx(5/3)^(2)` `=16.67J`

Discussion

No Comment Found

Related InterviewSolutions

A fly wheel rotating about a fixed axis has a kinetic energy of `360 J`. When its angular speed is `30 rad s^(-1)`. The moment of inertia of the wheel about the axis of rotation isA. `0.6kgm^(-2)`B. `0.15kgm^(-2)`C. `0.8kgm^(-2)`D. `0.75kgm^(-2)`
A constant external torque `tau` acts for a very brief period `/_ ` on a rotating system having moment of inertia `I` thenA. the angular velocity of the system will change by `tau/_ //I`B. the angular velocity of the system will change by `tau/_ //I`C. if the system was initially at rest, it will acquire rotational kinetic enegy `(tau/_ )^(2)//2I`D. the kinetic energy of the system will change by `(tau/_ )^(2)//(2I)`
A disc of mass `m` and radius `R` moves in the `x-y` plane as shown in Fig. The angular momentum of the disc about the origin `O` at the instant shown is A. `-5/2mR^(2)omegahatk`B. `7/3mR^(2)omegahatk`C. `-9/2mR^(2)omegahatk`D. `5/2mR^(2)omegahatk`
If `vec(tau)xxvec(L)=0` for a rigid body, where `vec(tau)=` resultant torque & `vec(L)` =angular momentum about a point and both are non-zero. ThenA. `vec(L)=`constantB. `|vec(L)|`=constantC. `|vec(L)|` will increasesD. `|vec(L)|` may increase
A particle has a linear momentum p and position vector r. the angular momentum of this particle about the origin will not be zero under the conditionsA. p=0B. p is perpendicular to r and does not cross rC. p is anti-parallel to rD. the particle passes through the origin.
The end B of the rod AB which makes angle `theta` with the floor is being pulled with a constant velocity `v_(v)` as shown. The length of the rod is `l`.A. `3/5v_(0)`B. `4/5v_(0)`C. `5/3v_(0)`D. `5/4v_(0)`
End `A` of a rod `AB` is being pulled on the floor with a constant velocity `v_(0)` as shown. Taking the length of the rod as `l`, at an instant when the rod makes an angle `37^@` with the horizontal, calculate the velocity of the `CM` of the rodA. `5/7v_(0)` at `tan^(-1)4/3` below horizontalB. `5/7v_(0)` at `tan^(-1)3/4` below horizontalC. `5/6v_(0)` at `tan^(-1)3/4` below horizontalD. `5/6v_(0)` at `tan^(-1) 4/3` below horizontal
A sphere of mass `M` and radius r shown in the figure on a rough horizontal plane. At some instant it has translational velocity `v_(0)`, and rotational velocity about centre `v_(0)//2r`. The percentage change of translational velocity after the sphere start pure rolling: A. `14.28%`B. `7.14%`C. `21.42%`D. none
A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.A. `1.2 rads^(-1)`B. `0.4rads^(-1)`C. `2.0rads^(-1)`D. `0.7rads^(-1)`
In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is smooth, what is the total kinetic energy of the body after `2s`?A. `25J`B. `16.67J`C. `50J`D. `37.5J`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment