Skip to main content
Pearson+ LogoPearson+ Logo
Ch 08: Dynamics II: Motion in a Plane
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 08: Dynamics II: Motion in a PlaneProblem 62a
Chapter 8, Problem 62a

2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ = 30°. What is the ball's speed when θ = 30°?

Verified step by step guidance
1
Step 1: Identify the forces acting on the ball at the given position (θ = 30°). These include the tension in the string (T = 20 N) and the gravitational force (F_g = m * g, where m = 2.0 kg and g = 9.8 m/s²). The net force provides the centripetal force required for circular motion.
Step 2: Break the gravitational force into components. The component of the gravitational force along the string is F_g_parallel = m * g * cos(θ), and the perpendicular component is F_g_perpendicular = m * g * sin(θ).
Step 3: Write the equation for the net force along the string, which provides the centripetal force: T - F_g_parallel = m * v² / r, where r is the radius of the circle (r = 0.80 m) and v is the speed of the ball.
Step 4: Substitute the known values into the equation: T - (m * g * cos(θ)) = m * v² / r. Rearrange the equation to solve for v²: v² = r * (T - m * g * cos(θ)) / m.
Step 5: Take the square root of both sides to find the speed v: v = √[r * (T - m * g * cos(θ)) / m]. Substitute the numerical values for r, T, m, g, and θ to calculate the speed.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force acting on an object moving in a circular path, directed towards the center of the circle. In this scenario, the tension in the string provides the necessary centripetal force to keep the ball moving in a vertical circle. The balance of forces, including gravitational force and tension, determines the net centripetal force acting on the ball.
Recommended video:
Guided course
06:48
Intro to Centripetal Forces

Newton's Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for calculating the ball's speed, as the net force acting on the ball at the angle θ must be analyzed to find the acceleration and subsequently the speed of the ball in circular motion.
Recommended video:
Guided course
06:54
Intro to Forces & Newton's Second Law

Kinematics in Circular Motion

Kinematics in circular motion involves the study of the motion of objects traveling along a circular path. The relationship between linear speed, radius, and angular position is essential for solving the problem. By applying the principles of circular motion, one can derive the ball's speed at a specific angle using the radius of the circle and the forces acting on the ball.
Recommended video:
Guided course
03:48
Intro to Circular Motion
Related Practice
Textbook Question

An airplane feels a lift force L→\overrightarrow{L} perpendicular to its wings. In level flight, the lift force points straight up and is equal in magnitude to the gravitational force on the plane. When an airplane turns, it banks by tilting its wings, as seen from behind, by an angle from horizontal. This causes the lift to have a radial component, similar to a car on a banked curve. If the lift had constant magnitude, the vertical component of L→\overrightarrow{L} would now be smaller than the gravitational force, and the plane would lose altitude while turning. However, you can assume that the pilot uses small adjustments to the plane's control surfaces so that the vertical component of L→\overrightarrow{L} continues to balance the gravitational force throughout the turn. Find an expression for the banking angle θ\theta needed to turn in a circle of radius rr while flying at constant speed vv.

1404
views
Textbook Question

A 100 g ball on a 60-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. What was the tension in the string an instant before it broke?

1103
views
Textbook Question

Scientists design a new particle accelerator in which protons (mass 1.7 X 10-27 kg) follow a circular trajectory given by r=ccos(kt2)i+csin(kt2)j\mathbf{r} = c \cos(kt^2) \mathbf{i} + c \sin(kt^2) \mathbf{j} where c = 5.0 m and k = 8.0 x 104 rad/s2 are constants and t is the elapsed time. What is the radius of the circle?

1199
views
Textbook Question

For safety, elevators have a rotational governor, a device that is attached to and rotates with one of the elevator's pulleys. The governor, shown in FIGURE P8.63, is a disk with two hollow channels holding springs with metal blocks of mass m attached to their free ends. The faster the governor spins, the more the springs stretch. At a critical angular velocity ωc, the metal blocks contact the housing, which completes a circuit and activates an emergency brake. The spring force on a mass, which we will explore more thoroughly in Chapter 9, is FSp = k(r - L), where k is the spring constant measured in N/m, and L is the relaxed (unstretched) length of the spring. Suppose a rotational governor has L = 0.80R and the emergency brake activates when the metal blocks reach r = R. What is the critical angular velocity in rpm if R = 15cm, k = 20 N/m, and m = 25g? Ignore gravity.

94
views
Textbook Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force F⃗wind=−Fwindı^.\vec{F}_{\text{wind}} = -F_{\text{wind}} \hat{\imath}.. Find an expression for the angle at which the range is maximum.

1573
views
Textbook Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force F⃗wind=−Fwindı^.\vec{F}_{\text{wind}} = -F_{\text{wind}} \hat{\imath}. By what percentage is the maximum range of a 0.50 kg ball reduced if Fwind=0.60 NF_{\text{wind}}=0.60\text{ N}?

1350
views