A wheel having moment of inertia 2 kg-m² about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in one minute would be
ω = ω0 + α.t
0= (60 × 2π /60) + α.60
⇒ α = - π/30
Torque = I α = (2× π/30) = (π/15) N-m
The figure shows a horizontal block of mass M suspended by two wires A and B. The centre of mass of the block is closer to B than A. (i) Is the magnitude of the torque due to wire A is greater, less or equal to that due to B w.r.t. centre of mass ? (ii) Which wire A or B exerts more force on the block ?
As the object is in rotational equilibrium, Net torque acting on the object is zero.
so, Torque of TA = Torque of TB
TAXA= TBXB
\[ \frac{T_{A}}{T_{B}}=\frac{X_{B}}{X_{A}} \]
\[ X_{B} < X_{A} \]
\[ T_{A} < T_{B} \]
A wire of mass m and length L is bent in the form of a circular ring. The moment of inertia of the ring about its axis is
Let the radius of Ring formed is R. Then
2πR= L ⇒ R = L/2π
Moment of Interia= m (L/2π)²= mL²/ 4π²
Three rings each of mass m and radius r are so placed that they touch each other. The radius of gyration of the system about the axis as shown in the figure is
Moment of Inertia of ring about it's diameter is mR²/2. ( Using Perpendicular axis theorem)
Moment of Inertia about tangential axis in plane of ring will be mR²/2 + mR²= 3/2 mR²
Total moment of inertia about the axis shown in figure
= (3/2 mR² )× 2 + 1/2 mR²= 7/2 mR²
Radius of Gyration is k then 3m×k²= 7/2 mR²
so,
\[ k= \sqrt{\frac{7}{6}}r \]
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is
\[ \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}= mgh =mg\frac{3v^{2}}{4g}= \frac{3}{4}mv^{2} \]
\[ \frac{1}{2}I\omega^{2}= \frac{1}{4}mv^{2} \]
Solving I = MR²/2 so, Object is a disc or hollow cylinder.
A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be
\[ mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}= \frac{1}{2}mv^{2}+ \frac{1}{2}(\frac{2}{5}mR^{2})\frac{v^{2}}{R^{2}} \]
Solving we get,
\[ v= \sqrt[]{\frac{10}{7}gh} \]
A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
Given, rotational kinetic energy is 40% of total energy. so,
\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]
Solving ,
\[ I = \frac{2}{5}mR^{2} \]
Object is Solid Sphere.
A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity . Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be
If We take Ring and 4 small blocks as one system net Torque will be zero, Using Principal of conservation of angular momentum.
\[ I_{1}\omega= \left( I_{1}+ I_{2} \right)\omega_{1} \]
\[ MR^{2}\omega= \left( MR^{2}+ 4mR^{2} \right)\omega_{1} \]
\[ \omega_{1}= \left( \frac{M}{M+4m} \right)\omega \]
A particle is projected with a speed v at 45° with the horizontal. The magnitude of angular
momentum of the projectile about the point of projection when the particle is at its maximum height h is
Angular Momentum = momentum × ( Perpendicular distance of momentum from axis of rotation )
Angular Momentum = mv cos (45º) × h = mvh/√2
A particle of mass m = 5 units is moving with a uniform speed v = 3√2 m in the XOY plane along the line Y = X + 4. The magnitude of the angular momentum about origin is
Distance of line
\[ ax+by+c=0 \] from point (x1,y1) is given by
\[ d = \left( \frac{ax_{1}+ by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right) \]
So, distance of direction of velocity from origin is d= 2√2
Angular momentum = Perpendicular distance of momentum × momentum = 2√2 × 5 ×3√2= 60 Unit