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Chapter 6: Problem 90

A mass \(m\) is supported by a massless string wound around a uniform hollow cylinder of mass \(m\) and radius \(R\). If the string does not slip on the cylinder, with what acceleration will the mass fall on release? (A) \(\frac{2 g}{3}\) (B) \(\frac{g}{2}\) (C) \(\frac{5 g}{6}\) (D) \(\mathrm{g}\)

Short Answer

Expert verified
The acceleration of the falling mass when released is (B) \(\frac{g}{2}\).

Step by step solution

01

Define the problem and draw a free body diagram

We have a hanging mass \(m\) connected to a string that is wound around a uniform hollow cylinder of mass \(m\) and radius \(R\). When we release the mass, it falls due to gravity, unwinding the string and causing the cylinder to rotate. First, we should draw a free-body diagram to represent the forces acting on both the mass and the cylinder. For the mass \(m\), there's gravitational force \(mg\) acting downward, tension \(T\) in the string acting upward, and acceleration \(a\) acting downward. For the cylinder, there's also tension \(T\) in the string acting along the radius and torque \(\tau\) acting around the center of the cylinder.
02

Apply Newton's second law of motion

Now, we can apply Newton's second law of motion to the linear motion of the falling mass and the angular motion of the rotating cylinder. For the falling mass, we have: \(m\vec{a} = \vec{T} - m\vec{g}\) For the rotating cylinder, we have: \(\vec{\tau} = I\vec{\alpha}\) Here, \(\tau\) is the torque acting on the cylinder, \(I\) is the moment of inertia of the cylinder, and \(\alpha\) is the angular acceleration of the cylinder.
03

Calculate torque and moment of inertia for the cylinder

We can calculate torque \(\tau\) as the product of force and radius: \(蟿 = T \cdot R\) To find the moment of inertia of the hollow cylinder, we use the formula for the moment of inertia of a thin cylindrical shell with mass \(M\) and radius \(R\): \(I = M R^2\) Since here the mass of the cylinder is \(m\), we can write: \(I = m R^2\)
04

Relate angular acceleration to linear acceleration

As the string unwinds without slipping, the linear acceleration \(a\) of the falling mass is related to the angular acceleration \(\alpha\) of the cylinder by: \(a = R\alpha\) We can now substitute this relation into the equation for torque: \(TR = \left(mR^2\right)\left(\frac{a}{R}\right)\) Simplify the equation: \(TR = maR\)
05

Solve for acceleration

Now, we can use the equations for the linear and angular motions to solve for the acceleration \(a\). From the linear motion equation, we have: \(T = m(g - a)\) Substitute this expression for \(T\) into the angular motion equation: \(m(g - a)R = maR\) Divide by \(mR\): \(g - a = a\) Now, solve for \(a\): \(a = \frac{g}{2}\)
06

Choose the correct answer

The correct answer is (B) \(\frac{g}{2}\), as this is the acceleration of the falling mass when released.

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Most popular questions from this chapter

A hollow sphere, ring, disc and solid sphere each of mass \(1 \mathrm{~kg}\) and radius \(1 \mathrm{~m}\) is released from rest on an identical inclined plane of inclination \(37^{\circ} . \tan 37^{\circ}=3 / 4\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ). The co-efficient of friction between body and surface is \(\mu\). Then match the column.(A) \(12\left|a_{c m}\right|\) (1) \(9 \mathrm{~g}\) (B) \(\left|a_{A}\right|\) (2) \(g\) (C) \(\left|\frac{I \alpha}{2}\right|\) (3) \(\frac{g}{3}\) \(8 g\)

(A) Particle moving with speed (1) Kinetic \(v_{0}\) strikes to a rod placed on energy will a smooth table and sticks change. to it. (B) A thin rod of mass \(m\) and (2) Momentum length \(\ell\) inclined at an angle is conserved. \(\theta\) with horizontal is dropped on a smooth horizontal plane without any angular velocity. Its tip does not rebound after impact. (C) A solid sphere of mass (3) Angular \(m\) and radius \(R\) is rolling momentum \(\begin{array}{ll}\text { with velocity } v_{0} \text { along a } & \text { is conserved }\end{array}\) horizontal plane. It suddenly about any encounters an obstacle. point. (D) Two cylinders of radii \(r_{1}\) and (4) Angular \(r_{2}\) rotating about their axis momentum with angular speed \(\omega_{1}\) and is conserved \(\omega_{2}\) moved closer to touch \(\quad\) just before each other keeping their and after axis parallel. Cylinders first impact only slip over each other at the about point contact point but slipping \(\quad\) of impact. ceases after some time due to friction.

Rate of change of angular momentum about point \(O\) is (A) \(m v l\) (B) zero (C) \(m g l \cos \theta\) (D) \(m g l \sin \theta\)

Consider a uniform square plate of side \(a\) and mass \(m\). The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is (A) \(\frac{5}{6} m a^{2}\) (B) \(\frac{1}{12} m a^{2}\) (C) \(\frac{7}{12} m a^{2}\) (D) \(\frac{2}{3} m a^{2}\)

A thick-walled hollow sphere has outer radius \(R\). It rolls down on an inclined plane without slipping and its speed at bottom is \(v_{0}\). Now the incline is waxed so that the friction becomes zero. The sphere is observed to slide down without rolling and the speed now is (5 \(v_{0} / 4\) ). The radius of gyration of the hollow sphere about the axis through its centre is \(\frac{n R}{4}\). Then the value of \(n\) is.
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