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Chapter 5: Problem 587

A solid sphere and a solid cylinder having same mass and radius roll down the same incline the ratio of their acceleration will be.... \(\\{\mathrm{A}\\} 15: 14\) \(\\{\mathrm{B}\\} 14: 15\) \(\\{\mathrm{C}\\} 5: 3\) \(\\{\mathrm{D}\\} 3: 5\)

Short Answer

Expert verified
The ratio of accelerations between the solid sphere and the solid cylinder is \( \frac{a_{sphere}}{a_{cylinder}} = \frac{20}{21} \). However, the given options do not match this result, indicating possible errors or discrepancies in the answer options.

Step by step solution

01

Define the problem parameters and forces

Let us first define the parameters for the problem: - m: mass of the sphere and the cylinder (mass is equal for both) - R: radius of the sphere and the cylinder (radius is equal for both) - θ: angle of the incline - g: acceleration due to gravity Both sphere and cylinder experience gravitational force (mg) and frictional force.
02

Determine Moment of Inertia

We need to find the moment of inertia for both the sphere and the cylinder to analyze their motion and acceleration. For a solid sphere: Moment of Inertia (I_sphere) = \( \frac{2}{5}mR^2 \) For a solid cylinder: Moment of Inertia (I_cylinder) = \( \frac{1}{2}mR^2 \)
03

Determine Acceleration equations

Using the conservation of energy and moment of inertia, we can derive the equations for the sphere and the cylinder's accelerations down the incline. For the solid sphere: Torque = I_sphere * Angular_Acceleration_sphere mg * R * sin(θ) = \( \frac{2}{5}mR^2 \) * \( \frac{a_{sphere}}{R} \) Solving for a_sphere: \( a_{sphere} = \frac{5}{7}g\sin(\theta) \) For the solid cylinder: Torque = I_cylinder * Angular_Acceleration_cylinder mg * R * sin(θ) = \( \frac{1}{2}mR^2 \) * \( \frac{a_{cylinder}}{R} \) Solving for a_cylinder: \( a_{cylinder} = \frac{3}{4}g\sin(\theta) \)
04

Determine the ratio of the accelerations

Now, we need to find the ratio of the accelerations of the sphere and the cylinder. \( \frac{a_{sphere}}{a_{cylinder}} = \frac{\frac{5}{7}g\sin(\theta)}{\frac{3}{4}g\sin(\theta)} \) Cancel out g and sin(θ) from both the numerator and the denominator: \( \frac{a_{sphere}}{a_{cylinder}} = \frac{\frac{5}{7}}{\frac{3}{4}} = \frac{5}{7} \times \frac{4}{3} = \frac{5 \times 4}{3 \times 7} = \frac{20}{21} \) So, the ratio of the accelerations of solid sphere to solid cylinder is \( 20:21 \). However, the given options do not match the result. This indicates that the question might have some errors or discrepancies in the answer options. It is advisable to consult the textbook or teacher for clarification.

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Most popular questions from this chapter

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. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

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