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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 body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

Statement \(-1\) -If the cylinder rolling with angular speed- w. suddenly breaks up in to two equal halves of the same radius. The angular speed of each piece becomes \(2 \mathrm{w}\). Statement \(-2\) - If no external torque outs, the angular momentum of the system is conserved. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \(\\{\mathrm{B}\\}\) Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1 .\) \(\\{\mathrm{C}\\}\) Statement \(-1\) is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true
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