A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of $60°$ and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of $37°$ with the vertical.
Download our app for free and get started
Let l = length of the rod, and m = mass of the rod. Applying energy principle $\Big(\frac{1}{2}\Big)\text{l}\omega^2-0=\text{mg}\Big(\frac{1}{2}\Big)(\cos37^\circ-\cos60^\circ)$
$\Rightarrow\frac{1}{2}\times\frac{\text{ml}^2}{3}\omega^2$
$=\text{mg}\times\frac{1}{2}\Big(\frac{4}{5}-\frac{1}{2}\Big)\text{t}$
$\Rightarrow\omega^2=\frac{9\text{g}}{10\text{l}}=0.9\Big(\frac{\text{g}}{\text{l}}\Big)$ Again $\Big(\frac{\text{ml}^2}{3}\Big)\alpha=\text{mg}\Big(\frac{1}{2}\Big)\sin37^\circ=\text{mgl}\times\frac{3}{5}$
$\therefore\alpha=0.9\Big(\frac{\text{g}}{\text{l}}\Big)=$ angular acceleration. So, to find out the force on the particle at the tip of the rod $F_i$ = centrifugal force $=(\text{dm})\omega^2\text{l}=0.9(\text{dm})\text{g}$
$F_t$ = tangential force $=(\text{dm})\alpha\text{l}=0.9(\text{dm})\text{g}$ So, total force $\text{F}=\sqrt{\big(\text{F}_{\text{i}}^2+\text{F}_{\text{t}}^2\big)}=0.9\sqrt2(\text{dm})\text{g}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1As shown in the two sides of a step ladder BA and CA are $1.6m$ long and hinged at A. A rope DE, $0.5m$ is tied half way up. A weight $40kg$ is suspended from a point F, 1.2m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take $g = 9.8m/s^2$) (Hint: Consider the equilibrium of each side of the ladder separately).View Solution
- 2View SolutionA wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre.How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?
- 3The oxygen molecule has a mass of $5.30 \times 10^{-26}kg$ and a moment of inertia of $1.94 \times 10^{-46}kgm^2$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500m/s$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.View Solution
- 4Locate the centre of mass of a system of particles of mass $m_1 = 1kg, m_2 = 2kg$ and $m_3 = 3kg$, situated at the corners of an equilateral triangle of side 1.0 metre.View Solution
- 5As shown in the two sides of a step ladder BA and CA are 1.6m long and hinged at A. A rope DE, 0.5m is tied half way up. A weight 40kg is suspended from a point F, 1.2m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take $g = 9.8m/s^2$) (Hint: Consider the equilibrium of each side of the ladder separately).View Solution
- 6View SolutionA particle of mass m is projected from origin O with speed u at an angle with positive x-axis. Find the angular momentum of particle at any time t about O before the particle strikes the ground again.
- 7View SolutionA solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
- Will it reach the bottom with the same speed in each case?
- Will it take longer to roll down one plane than the other?
- If so, which one and why?
- 8View SolutionTwo particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
- 9View SolutionSuppose the particle of the previous problem has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M:
- Find the velocity of the centre of mass C of the system constituting ''The rod plus the particle''.
- Find the velocity of the particle with respect to C before the collision.
- Find the velocity of the rod with respect to C before the collision.
- Find the angular momentum of the particle and of the rod about the centre of mass C before the collision.
- Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision.
- Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.
- 10View SolutionFig. shows two blocks of masses 5kg and 2kg placed on a frictionless surface and connected by a spring. An external kick gives a velocity 14 m/ s to the heavier block in the direction of lighter one. Deduce (i) the velocity gained by the centre of mass and (ii) the separate velocities of the two blocks in the centre of mass coordinates just after the kick.