Estimate the moment of inertia of a bicycle wheel 67 cm in diameter. The rim and tire have a combined mass of 1.1 kg. The mass of the hub (at the center) can be ignored (why?).

Short Answer

Expert verified

The moment of inertia of the wheel is \(0.12\;{\rm{kg}}\;{{\rm{m}}^2}\).

Step by step solution

01

Identification of the given data

The mass of the wheel is \(m = 1.1\;{\rm{kg}}\).

The diameter of the wheel is \(d = 67\;{\rm{cm}} = 0.67\;{\rm{m}}\).

02

Definition of moment of inertia

Moment of inertia is a quantity that expresses a body’s tendency to resist angular acceleration about the axis of rotation.

It is given as the product of the mass of a rigid body and the square of the distance from the axis of rotation.

\(I = m{r^2}\)

03

Determination of the moment of inertia of the solid sphere

Consider the rim and the tire as a hoop where all of the significant masses are located at the same distance \(\left( {r = \frac{d}{2}} \right)\) from the axis of rotation. Thus, the moment of inertia of the wheel will be given by:

\(\begin{align}I &= m{r^2}\\ &= \left( {1.1\;{\rm{kg}}} \right){\left( {\frac{{0.067}}{2}\;{\rm{m}}} \right)^2}\\ &= \left( {1.1\;{\rm{kg}}} \right){\left( {0.335\;{\rm{m}}} \right)^2}\\ &= 0.12\;{\rm{kg}}\;{{\rm{m}}^2}\end{align}\)

The mass of the hub can be ignored because its distance from the axis of rotation is significantly less (approximately r = 0). So, it has a very small rotational inertia.

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Most popular questions from this chapter

A cyclist accelerates from rest at a rate of \({\bf{1}}{\bf{.00}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). How fast will a point at the top of the rim of the tire (diameter = 0.80 cm) be moving after 2.25 s? [Hint: At any moment, the lowest point on the tire is in contact with the ground and is at rest — sees Fig. 8–57.]

FIGURE 8-57 Problem 79

Calculate the moment of inertia of an align of point objects, as shown in Fig. 8–47 about (a) the y axis and (b) the x-axis. Assume two masses, \(m = 2.2\;{\rm{kg}}\)and \(M = 3.4\;{\rm{kg}}\), and the objects are wired together by very light, rigid pieces of wire. The align is rectangular and split through the middle by the x-axis. (c) About which axis would it be harder to accelerate this align?

FIGURE 8-47

Problem 39

Assume that a 1.00-kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle, as shown in Fig. 8–46. The ball is accelerated uniformly from rest to 8.5 m/s in 0.38 s, at which point it is released. Calculate (a) the angular acceleration of the arm and (b) the force required for the triceps muscle. Assume that the forearm has a mass of 3.7 kg, and it rotates like a uniform rod about an axis at its end.

FIGURE 8-46

Problems 35 and 36

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius \({{\bf{R}}_{\bf{1}}}{\bf{ = 2}}{\bf{.5}}\;{\bf{cm}}\) and winds its way out to radius \({{\bf{R}}_{\bf{2}}}{\bf{ = 5}}{\bf{.8}}\;{\bf{cm}}\). To read the digital information, a CD player rotates the CD so that the player’s readout laser scans along the spiral’s sequence of bits at a constant linear speed of 1.25 m/s. Thus the player must accurately adjust the rotational frequency f of the CD as the laser moves outward. Determine the values for f (in units of rpm) when the laser is located at \({{\bf{R}}_{\bf{1}}}\) and when it is at \({{\bf{R}}_{\bf{2}}}\).

Question:(II) Two masses,\({m_{\bf{A}}} = 32.0 kg\)and\({m_{\bf{B}}} = 38.0 kg\)are connected by a rope that hangs over a pulley (as in Fig. 8–54). The pulley is a uniform cylinder of radius\(R = 0.311 m\)and mass 3.1 kg. Initially\({m_{\bf{A}}}\)is on the ground and\({m_{\bf{B}}}\)rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of\({m_{\bf{B}}}\)just before it strikes the ground. Assume the pulley bearing is frictionless.

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