A uniform rod of length L rests against a smooth roller as shown in figure. Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is $\theta.$ 
Download our app for free and get started
Rod has a length = LIt makes an angle $\theta$ with the floor
The vertical wall has a height = h
$\text{R}_2=\text{mg}-\text{R}_1\cos\theta\ \dots(1)$
$\text{R}_1\sin\theta=\mu\text{R}_2\ \dots(2)$
$\text{R}_1\cos\theta\times\Big(\frac{\text{h}}{\tan\theta}\Big)+\text{R}_1\sin\theta\times\text{h}=\text{mg}\times\frac{1}{2}\cos\theta$
$\Rightarrow\text{R}_1\times\Big(\frac{\cos^2\theta}{\tan\theta}\Big)+\text{R}_1\sin\theta\times\text{h}=\text{mg}\times\frac{1}{2}\cos\theta$
$\Rightarrow\text{R}_1=\frac{\text{mg}\times\frac{\text{L}}{2\cos\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}}$
$\Rightarrow\text{R}_1\cos\theta=\frac{\frac{\text{mg}\text{L}}{2\cos^2\theta\sin\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}}$
$\Rightarrow\mu=\frac{\text{R}_1\sin\theta}{\text{R}_2}=\frac{\frac{\text{mgL}}{2\cos\theta\sin\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}\text{mg}-\text{mg}\frac{1}{2}\cos^2\theta}$
$\Rightarrow\mu=\frac{\frac{\text{L}}{2\cos\theta.\sin\theta}\times2\sin\theta}{2\big(\cos^2\theta\text{h}+\sin^2\theta\text{h}\big)-\text{L}\cos^2\theta\sin\theta}$
$\Rightarrow\mu=\frac{\text{L}\cos\theta\sin^2\theta}{\text{2h}-\text{L}\cos^2\theta\sin\theta}$
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
- 1A disc of radius R is rotating with an angular speed $\omega_\text{o}$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu_\text{k}{:}$View Solution
- What was the velocity of its centre of mass before being brought in contact with the table?
- What happens to the linear velocity of a point on its rim when placed in contact with the table?
- What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
- Which force is responsible for the effects in (b) and (c).
- What condition should be satisfied for rolling to begin?
- Calculate the time taken for the rolling to begin.
- 2View Solution
- Prove the theorem of perpendicular axes.
- Prove the theorem of parallel axes.
- 3View 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.
- 4View SolutionA body is in translational equilibrium under the action of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?
- 5View SolutionEstablish the relationship between Torque and Moment of Inertia.
- 6View Solution
- Prove the theorem of perpendicular axes.
- Prove the theorem of parallel axes.
- 7View SolutionFour bodies have been arranged at the corners of a rectangle shown in figure. Find the centre of mass of the system.
- 8A car weighs $1800kg$. The distance between its front and back axles is $1.8m$. Its centre of gravity is $1.05m$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.View Solution
- 9From a uniform disk of radius R, a circular hole of radius $\frac{\text{R}}{2}$ is cut out. The centre of the hole is at $\frac{\text{R}}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.View Solution
- 10View SolutionA uniform disc of radius R and mass M is mounted on an axis supported in fixed frictionless bearing. A light chord is wrapped around the rim of the wheel and suppose that we hang a body of mass m from the chord. Find the angular acceleration of the disc and tangential acceleration of point on the rim.