/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Block \(A\) rests on a horizonta... [FREE SOLUTION] | 91Ó°ÊÓ

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Block \(A\) rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block \(B\) is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius \(0.080 \mathrm{~m}\) ) because the pulley rotates on a frictionless axle. The horizontal surface on which block \(A\) (mass \(2.50 \mathrm{~kg}\) ) moves is frictionless. The system is released from rest, and block \(B\) (mass \(6.00 \mathrm{~kg}\) ) moves downward \(1.80 \mathrm{~m}\) in \(2.00 \mathrm{~s}\). (a) What is the tension force that the rope exerts on block \(B ?\) (b) What is the tension force on block \(A ?\) (c) What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Short Answer

Expert verified
The tension force that the rope exerts on block B is \(55.3 N\), the tension force on block A is \(1.13 N\), and the moment of inertia of the pulley is \(9.63 kg*m^2\).

Step by step solution

01

Identify known quantities

Block B is moving downwards under the influence of the gravitational force. Its mass is 6.00 kg, the displacement is 1.80 m and the time taken is 2.00 s. Block A, of mass 2.50 kg, is stationary on a frictionless surface. The pulley has a radius of 0.080 m.
02

Apply second law of motion to Block B

The equation of motion for Block B is \(F=ma\). The downward force is the weight \(F_g=m_Bg=6.00 kg*9.8 m/s^2 = 58.8 N\). The acceleration \(a\) is the change in velocity \(v\) over time, and velocity is the change in displacement over time, so \(a=\Delta x / \Delta t^2=1.80 m/ (2.00 s)^2 =0.45 m/s^2\). So now the tension force \(T_B\) on block B which is \(T_B= m_Bg - m_Ba = 58.8 N - 6.00 kg*0.45 m/s^2 = 55.3 N\).
03

Apply second law of motion to Block A

In the horizontal direction for Block A, the only force acting is the tension, since there is no friction. So, \(F_{net} = m_Aa\) which means that the tension force \(T_A\) on block A is given by \(T_A = m_Aa = 2.50 kg*0.45 m/s^2 = 1.13 N\).
04

Calculate the moment of inertia for the Pulley

The tension in the string exerts a torque on the pulley, causing it to rotate. The torques caused by \(T_A\) and \(T_B\) are equal and opposite. From \(\tau = I \alpha = r F_{net}\), one can find the moment of inertia \(I\). The net force here is \(F_{net} = T_B - T_A = 55.3 N - 1.13 N = 54.17 N\). Thus, \(I = \frac {r F_{net}}{\alpha} = 0.080 m*54.17 N/0.45 m/s^2 = 9.63 kg*m^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law of Motion
Understanding this law is crucial for solving a variety of physics problems, including tension force scenarios like in the textbook exercise. Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The formula can be expressed as \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

In the given exercise, we apply this law twice: once for Block B, which is moving downwards due to gravity minus the tension in the rope, and once for Block A, which is being pulled horizontally by the same tension. Independent of its direction, the tension force creates an acceleration in Block A that is equal in magnitude but opposite in direction to the acceleration of Block B, as dictated by Newton's Third Law. Thus, by calculating the acceleration of Block B, we indirectly determine the tension force in Block A.
Rotational Motion and Moment of Inertia
Rotational motion involves objects that rotate about an axis, and the moment of inertia \( I \) is a measure of an object's resistance to change in its rotational motion. It's analogous to mass in linear motion. The moment of inertia depends on the mass distribution relative to the rotation axis. For standard shapes, like cylinders, spheres, and discs, there are established formulas to calculate it.

The pulley in the exercise represents a real-world application of rotational motion. It has a moment of inertia that needs to be overcome by the torque generated by the tension forces in the rope. Torque \( \tau \) is calculated by the product of the force and the perpendicular distance from the axis of rotation \( r \) expressed as \( \tau = r F_{net} \). To find the moment of inertia for the pulley, a rearranged version of the torque equation \( \tau = I \alpha \) is used, which combines rotational motion with Newton’s Second Law, where \( \alpha \) is the angular acceleration.
Dynamics of Rigid Bodies
The dynamics of rigid bodies involves analyzing the movement of objects that do not deform under the influence of forces. A rigid body can move translationally, as in the horizontal movement of Block A, and rotationally, like the pulley. The complexities arise when these two types of motion are interlinked, as the rope's tension creates a translational motion in Block A and a rotational motion in the pulley.

It’s vital to consider the moment of inertia for rotational motion, similar to how mass is involved in translational motion. The dynamics of the rigid body (the pulley) in motion becomes a significant part of the analysis. The calculation of the moment of inertia in the last step of the solution considers the pulley as a rigid rotating body. The dynamics of rigid bodies are governed by Newton's laws too, but with torque and moment of inertia as key factors in rotational motion.

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Most popular questions from this chapter

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\), connected by a light axle of radius \(b\). A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

Example 10.7 discusses a uniform solid sphere rolling with- out slipping down a ramp that is at an angle \(\beta\) above the horizontal. Now consider the same sphere rolling without slipping up the ramp. (a) In terms of \(g\) and \(\beta\), calculate the acceleration of the center of mass of the sphere. Is your result larger or smaller than the acceleration when the sphere rolls down the ramp, or is it the same? (b) Calculate the friction force (in terms of \(M, g,\) and \(\beta\) ) for the sphere to roll without slipping as it moves up the incline. Is the result larger, smaller, or the same as the friction force required to prevent slipping as the sphere rolls down the incline?

A hollow, thin-walled sphere of mass \(12.0 \mathrm{~kg}\) and diameter \(48.0 \mathrm{~cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time \(3.00 \mathrm{~s}\), find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

A thin uniform rod has a length of \(0.500 \mathrm{~m}\) and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of \(0.400 \mathrm{rad} / \mathrm{s}\) and a moment of inertia about the axis of \(3.00 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\). A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is \(0.160 \mathrm{~m} / \mathrm{s}\). The bug can be treated as a point mass. What is the mass of (a) the rod; (b) the bug?

A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle \(\beta\) to the horizontal (see Example 10.7 in Section 10.3 ). Treat the ball as a uniform solid sphere, ignoring the finger holes. (a) Draw the free-body diagram for the ball. Explain why the friction force must be directed uphill. (b) What is the acceleration of the center of mass of the ball? (c) What minimum coefficient of static friction is needed to prevent slipping?

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