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
What is moment of inertia in terms of angular momentum (L) and kinetic energy (K)
\[ L = I \omega \]
and
\[ K = \frac{1}{2} I \omega^{2} \]
Squaring L and Dividing it with K we get,
\[ I= \frac{L^{2}}{2K} \]
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
A conical pendulum consists of a simple pendulum moving in a horizontal circle as shown. C is the pivot, O the centre of the circle in which the pendulum bob moves and ω the constant angular velocity of the bob. If L is the angular momentum about point C, then
As object is in circular motion angular momentum
\[ \vec L=I\vec\omega \]
Direction of omega is along the axis so, L will have direction along axis OC. So both magnitude
and direction of angular momentum L is constant.
A uniform rectangular plate of mass m which is free to rotate about the smooth vertical hinge passing through the centre and perpendicular to the plate, is lying on a smooth horizontal surface. A particle of mass m moving with speed ‘u’ collides with the plate and sticks to it as shown in figure. The angular velocity of the plate after collision will be
Taking rectangle and the object as one system angular momentum is conserved.
so, mva= ( m(√5a/2)² + m((2a)²+ a²)/12) ω
\[ \omega = \frac{3u}{5a} \]
A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin O is
For Rolling L = MvR + Iω = M(ωR)R + (MR²/2)ω = (3/2)MR²ω