21.
A particle of mass $$m$$ rotates with a uniform angular speed $$\omega .$$ It is viewed from a frame rotating about the $$z$$-axis with a uniform angular velocity $${\omega _0}.$$ The centrifugal force on the particle is :
A
$$m{\omega ^2}r$$
B
$$m\omega _0^2r$$
C
$$m\left( {\frac{{\omega + {\omega _0}}}{2}} \right)a$$
The centrifugal force on the particle $$ = {\text{mass}} \times {\text{acceleration of the frame}}$$
$$\eqalign{
& = m \times \omega _0^2r \cr
& = m\omega _0^2r \cr} $$
22.
One end of the string of length $$l$$ is connected to a particle of mass $$m$$ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed $$v,$$ the net force on the particle (directed towards center) will be ($$T$$ represents the tension in the string)
Consider the string of length $$l$$ connected to a particle as shown in the figure.
Speed of the particle is $$v.$$ As the particle is in uniform circular motion, net force on the particle must be equal to centripetal force which is provided by the tension $$\left( T \right).$$
$$\therefore {\text{Net}}\,{\text{force}} = {\text{Centripetal}}\,{\text{force}} \Rightarrow \frac{{m{v^2}}}{l} = T$$
23.
A uniform rod $$AB$$ of length $$3r$$ remains in equilibrium on a hemispherical bowl of radius $$r$$ as shown in figure. Ignoring friction, the inclination of the rod $$\theta $$ with the horizontal is
From the geometry
$$\eqalign{
& AD = 2r;AC = 2r\cos \theta \cr
& CD = 2r\sin \theta \cr
& AG = 1.5r;GC = \left( {2r\cos - 1.5r} \right) \cr
& \tan \theta = \frac{{GC}}{{CD}} = \frac{{2r\cos \theta - 1.5r}}{{2r\sin \theta }} \cr} $$
After simplifying, we get, $$\cos \theta = 0.9.$$
24.
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum friction of the length of the chain that can hang over one edge of the table is
The force of friction should balance the weight of chain hanging. If $$M$$ is the mass of whole chain of length $$L$$ and $$x$$ is the length of chain hanging to balance, then
$$\eqalign{
& \mu \frac{M}{L}\left( {L - x} \right)g = \frac{M}{L}xg \cr
& {\text{or}}\,\mu \left( {L - x} \right) = x \cr
& {\text{or}}\,x = \frac{{\mu L}}{{\mu + 1}} = \frac{{0.25L}}{{1.25}}\,\,{\text{ }}\left( {As,\mu = 0.25} \right) \cr
& \therefore x = \frac{L}{5} \cr
& {\text{or}}\,\frac{x}{L} = \frac{1}{5} = \frac{1}{5} \times 100 = 20\% \cr} $$
25.
A mass $$'m'$$ is supported by a massless string wound around a uniform hollow cylinder of mass $$m$$ and radius $$R.$$ If the string does not slip on the cylinder, with what acceleration will the mass fall or release?
From figure,
Acceleration $$a = R\alpha \,......\left( {\text{i}} \right)$$
and $$mg - T = ma\,......\left( {{\text{ii}}} \right)$$
From equation (i) and (ii)
$$\eqalign{
& T \times R = m{R^2}\alpha = m{R^2}\left( {\frac{a}{R}} \right) \cr
& {\text{or}}\,T = ma \cr
& \Rightarrow mg - ma = ma \cr
& \Rightarrow a = \frac{g}{2} \cr} $$
26.
A particle is moving with a uniform speed in a circular orbit of radius $$R$$ in a central force inversely proportional to the $${n^{th}}$$ power of $$R.$$ If the period of rotation of the particle is $$T,$$ then:
A
$$T \propto {R^{\frac{3}{2}}}\,{\text{for}}\,{\text{any}}\,n.$$
B
$$T \propto {R^{\frac{n}{{2 + 1}}}}$$
C
$$T \propto {R^{\frac{{\left( {n + 1} \right)}}{2}}}$$