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Chapter 9: Problem 37

A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Short Answer

Expert verified
The tangential velocity is approximately 8.21 m/s.

Step by step solution

01

Understanding the Relationship between Kinetic Energy and Moment of Inertia

The kinetic energy \( K \) of a rotating object is given by the formula: \( K = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. For a solid sphere rotating about its diameter, \( I = \frac{2}{5} m r^2 \), where \( m \) is mass and \( r \) is the radius.
02

Solving for Angular Velocity \( \omega \)

Given \( K = 236 \text{ J} \), \( m = 28.0 \text{ kg} \), and \( r = 0.380 \text{ m} \), we substitute the moment of inertia into the kinetic energy formula: \( 236 = \frac{1}{2} \left(\frac{2}{5} \cdot 28.0 \cdot 0.380^2\right) \omega^2 \). First, calculate the moment of inertia: \( I = \frac{2}{5} \cdot 28.0 \cdot 0.380^2 = 1.01232 \text{ kg} \cdot \text{m}^2 \). Substitute back to find \( \omega \): \( 236 = \frac{1}{2} \times 1.01232 \times \omega^2 \). Solving for \( \omega \), we find \( \omega \approx 21.61 \text{ rad/s} \).
03

Calculating Tangential Velocity

Once \( \omega \) is known, the tangential velocity \( v_t \) of a point at the edge of the sphere is related by the formula \( v_t = \omega r \). Substituting in the known values: \( v_t = 21.61 \cdot 0.380 = 8.2118 \text{ m/s} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy in rotational dynamics refers to the energy an object possesses due to its rotation. Unlike linear kinetic energy, which depends on an object鈥檚 mass and velocity, rotational kinetic energy depends on the object's moment of inertia and angular velocity.
The formula used is \[ K = \frac{1}{2} I \omega^2 \] where:
  • \( K \) is the kinetic energy.
  • \( I \) is the moment of inertia, which captures how mass is distributed relative to the axis of rotation.
  • \( \omega \) represents the angular velocity, or how fast the object is rotating.
Understanding this relationship helps calculate how much energy is stored in a rotating system鈥攌ey for deciphering other dynamic properties like the sphere in our example.
Moment of Inertia
The moment of inertia is a measure that reflects how mass is positioned relative to the axis of rotation. Think of it as the rotational equivalent of mass in linear motion. It's crucial because it influences how easily an object can be put into rotational motion or stopped.
For different shapes and axes, the moment of inertia will vary. Focusing on a solid sphere rotating about its diameter, the moment of inertia is given by \[ I = \frac{2}{5} m r^2 \] where:
  • \( m \) is the mass of the sphere.
  • \( r \) is the radius of the sphere.
This formula helps describe how the mass distribution affects the sphere鈥檚 rotational characteristics, which is evident in our example where we calculated \( I \approx 1.01232 \text{ kg} \cdot \text{m}^2 \).
Angular Velocity
Angular velocity represents how quickly an object rotates around an axis. It is expressed in radians per second (\( \text{rad/s} \)), making it essential for understanding rotational motion just as regular velocity describes straight-line motion.
  • It gives insight into the speed of rotation.
  • Helps in determining other dynamic variables like tangential velocity.
From the kinetic energy equation \( K = \frac{1}{2} I \omega^2 \), solving for \( \omega \) helps us determine the rate at which the sphere spins. For our sphere example, we found \( \omega \approx 21.61 \text{ rad/s} \), indicating a fairly rapid spin around its axis.
Tangential Velocity
Tangential velocity refers to the linear speed of a point on the outer surface of a rotating object. It's what you would perceive as speed if you were standing on that edge point as the sphere spins.
Calculating tangential velocity involves multiplying angular velocity by the radius of the rotation: \[ v_t = \omega r \].This formula shows the direct relationship between how fast the sphere spins and how large its radius is.
  • \( v_t \) provides a more intuitive understanding of motion, as it's a linear speed.
  • Helps connect rotational dynamics with linear motion concepts.
In our scenario, we calculated \( v_t \approx 8.2118 \, \text{m/s} \), illustrating how quickly a point on the sphere's edge moves.

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Most popular questions from this chapter

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s\(^2\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{rad} = \omega^2r\) and (b) \(a_{rad} = v^2/r\)

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriendly dog is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof. If the wheel has radius 0.300 m and a moment of inertia of 9.60 kg \(\cdot\) m\(^2\) for rotation about the axle, how long does it take her to reach the sidewalk, and how fast will she be moving just before she lands? Ignore friction.

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm 1rev/min2. (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s\(^2\), what is its angular velocity at \(t =\) 2.50 s? (b) Through what angle has the wheel turned between \(t =\) 0 and \(t =\) 2.50 s?

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant \(linear\) speed of \(v =\) 1.25m/s. Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise 9.20.) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta) = r_0 + \beta\theta\), where \(r_0\) is the radius of the spiral at \(\theta =\) 0 and \(\beta\) is a constant. On a CD, \(r_0\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d\theta\), the distance scanned along the track is \(ds = rd\theta\). Using the above expression for \(r(\theta)\), integrate \(ds\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) Since the track is scanned at a constant linear speed \(v\), the distance s found in part (a) is equal to \(vt\). Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta\) ; choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_z\) and the angular acceleration \(\alpha_z\) as functions of time. Is \(\alpha_z\) constant? (d) On a CD, the inner radius of the track is 25.0 mm, the track radius increases by 1.55 mm per revolution, and the playing time is 74.0 min. Find \(r_0, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_z\) (in rad/s) versus \(t\) and \(\alpha_z\) (in rad/s\(^2\)) versus \(t\) between \(t =\) 0 and \(t =\) 74.0 min. \(\textbf{The Spinning eel.}\) American eels (\(Anguilla\) \(rostrata\)) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could.
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