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Chapter 5: Problem 642

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None

Short Answer

Expert verified
The frictional force on the disc rolling without slipping down an inclined plane at an angle θ with the horizontal is given by \(f = \frac{2Mg\sin\theta}{3}\), which corresponds to option B.

Step by step solution

01

Draw a free body diagram

Draw a free body diagram for the disc while it's rolling down the incline showing all the forces acting on it: gravitational force (Mg), frictional force (f), and normal force (N).
02

Determine the equations for the horizontal and vertical forces

Resolve the forces in the horizontal (x) and vertical (y) directions, and describe the acceleration and torque acting on the disc: Sum of the forces in the x-direction: \(f - Mg\sin\theta = Ma\) Sum of the forces in the y-direction: \(N - Mg\cos\theta = 0\)
03

Determine the rotational equation

Since the disc is rolling without slipping, we need to consider a rotational equation using the equation of torque about the center of the disc: Torque = I * angular acceleration \(fR = \frac{1}{2}MR^2α\) Where I is the moment of inertia of the disc (\(\frac{1}{2}MR^2\)) and α is the angular acceleration.
04

Determine the No-slip condition equation

A no-slip condition means that when the disc is rolling, the point of contact with the inclined plane has zero velocity. Using the relationship between linear acceleration and angular acceleration: \( a = Rα \)
05

Solve the equations for frictional force

Using Steps 2, 3, and 4, we can solve the system of equations to find the frictional force (f): From Step 4, \(α = \frac{a}{R}\). Plugging this value into Step 3 equation: \(fR = \frac{1}{2}MR^2 (\frac{a}{R}) \) Solving for f: \(f = \frac{1}{2}Ma\) Now, lets substitute \(f = \frac{1}{2}Ma\) into the Step 2 equation of horizontal forces: \(\frac{1}{2}Ma - Mg\sin\theta = Ma\) Solving for a: \(a = \frac{2}{3}g\sin\theta\) Finally, substituting the value of a back to frictional force equation: \(f = \frac{1}{2}M(\frac{2}{3}g\sin\theta)\)
06

Choose the correct answer

Comparing this result with the given options: \(f = \frac{2Mg\sin\theta}{3}\), which corresponds to option B.

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Most popular questions from this chapter

Two blocks of masses \(10 \mathrm{~kg}\) an \(4 \mathrm{~kg}\) are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of the lighter block. The velocity of the centre of mass is : \(\\{\mathrm{A}\\} 30 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{B}\\} 20 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{C}\\} 10 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{D}\\} 5 \mathrm{~m} / \mathrm{s}\)

A small object of uniform density rolls up a curved surface with initial velocity 'u'. It reaches up to maximum height of $3 \mathrm{v}^{2} / 4 \mathrm{~g}$ with respect to initial position then the object is \(\\{\mathrm{A}\\}\) ring \(\\{B\\}\) solid sphere \(\\{\mathrm{C}\\}\) disc \\{D\\} hollow sphere

A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm} .\) is removed from tve \(\mathrm{x}\) edge of the plate. Find the position of centre of mass of the remaining portion with respect to centre of mass of whole plate. $\\{\mathrm{A}\\}-7 \mathrm{~cm} \quad\\{\mathrm{~B}\\}+9 \mathrm{~cm} \quad\\{\mathrm{C}\\}-9 \mathrm{~cm} \quad\\{\mathrm{D}\\}+7 \mathrm{~cm}$

If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(\mathrm{T}\) is..... \(\\{\mathrm{A}\\}\left(\pi \mathrm{MR}^{3} / \mathrm{T}\right)\) \(\\{\mathrm{B}\\}\left(\operatorname{MR}^{2} \pi / \mathrm{T}\right)\) \(\\{C\\}\left(2 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\) \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\)

Two identical hollow spheres of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)
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