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Chapter 5: Problem 641

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 acceleration of the centre of mass of the disc is \(\\{\mathrm{A}\\} \mathrm{g} \sin \theta\) \(\\{B\\}[(2 g \sin \theta) / 3]\) \(\\{\mathrm{C}\\}[(\mathrm{g} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(2 \mathrm{~g} \cos \theta) / 3]\)

Short Answer

Expert verified
The acceleration of the center of mass of the uniform disc rolling without slipping down the inclined plane is given by \(a = \dfrac{2Mg\sin(\theta)}{3}\). Thus, the correct answer is option B.

Step by step solution

01

Setup the problem

Draw a diagram of the situation to visualize the forces acting on the disc. The forces acting on the disc are: the gravitational force (mg), the normal force (N), and the frictional force (f).
02

Using Newton's second law of motion

Apply Newton's second law (F = ma) to the forces along the inclined plane and perpendicular to the inclined plane. This divides the force into two different components: one parallel to the inclined plane, and one perpendicular to the inclined plane. (a) Along the inclined plane: \(F = M*a = Mgsin(\theta) - f \) (b) Perpendicular to the inclined plane: \(0 = N - Mgcos(\theta) \)
03

Torque and Rolling Condition

Since the disc rolls without slipping, the rolling condition relates the linear acceleration (a) with the angular acceleration (α) as follows: \(a = R*\alpha\). Now, calculate the torque (τ) acting on the disc. The only force contributing to the torque is friction (f) at the point of contact. \(τ =I*\alpha = f * R \)
04

Moment of Inertia

Calculate the moment of inertia (I) of the disc, as it is necessary to determine the torque acting on it. The moment of inertia for a uniform disc is given by: \(I = \dfrac{1}{2}MR^2 \)
05

Plug in and Solve

Now, plug in the values from Step 2, Step 3, and Step 4 to solve for the acceleration (a). With some algebra, we have: \(a = \dfrac{Mgsin(\theta) - f}{M} \) (from step 2a) \(f = I * \dfrac{a}{R} \) (from step 3) Substituting f and I and simplifying, we get: \(a = M*g*sin(\theta) - \dfrac{1}{2}*MR^2* \dfrac{a}{R}\) \(a = \dfrac{2Mgsin(\theta)}{3} \) Thus, the acceleration of the center of mass of the disc is: \(a = \dfrac{2Mgsin(\theta)}{3} \) So, the correct answer is the option B.

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Most popular questions from this chapter

A meter stick of mass \(400 \mathrm{gm}\) is pivoted at one end and displaced through an angle 600 the increase in its P.E. is \(\overline{\\{\mathrm{A}\\} 2}\) \(\\{B\\} 3\) \(\\{\) C \(\\}\) Zero \(\\{\mathrm{D}\\} 1\)

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