(II) A bowling ball of mass 7.25 kg and radius 10.8 cm rolls without slipping down a lane at\({\bf{3}}{\bf{.10 m/s}}\).Calculate its total kinetic energy.

Short Answer

Expert verified

The total kinetic energy of the bowling ball is 48.77 J.

Step by step solution

01

Identification of given data 

The given data can be listed below as:

  • The mass of the bowling ball is\(m = 7.25{\rm{ kg}}\).
  • The radius of the bowling ball is\(R = 10.8{\rm{ cm}}\).
  • The velocity of the bowling ball is \(v = 3.10{\bf{ }}{\rm{m/s}}\).
02

Understanding the motion of the bowling ball on a lane

The bowling ball is rolling from the top of the lane to its bottom.The kinetic energy of the bowling ball is equal to thesum of the rotational and translational kinetic energies.The rotational kinetic energy of the bowling ball can be expressed in terms of Joules.

03

Determination of the rotational and the translational kinetic energies of the bowling ball

The rotational kinetic energy of the can be expressed as:

\(K = \frac{1}{2}I{\omega ^2}\) … (i)

Here,\(\omega \)is the angular speed of the bowling ball.

The moment of inertia of the bowling ball can be expressed as:

\(I = \frac{2}{5}M{R^2}\)

Substitute the values in equation (i).

\(\begin{align}K &= \frac{1}{2} \times \frac{2}{5}M{R^2} \times {\omega ^2}\\K &= \frac{1}{5}M\left( {{R^2}{\omega ^2}} \right)\\K &= \frac{1}{5}M{v^2}\end{align}\)

\(K = \frac{1}{5}M{v^2}\) … (ii)

Here,\(v\)is the translational velocity of the bowling ball.

The translational kinetic energy of the bowling ball can be expressed as:

\({K_1} = \frac{1}{2}M{v^2}\) … (iii)

04

Determination of the total kinetic energy of the bowling ball

The total kinetic energy is equal to the sum of translational and rotational kinetic energies. Add equations (ii) and (iii) to obtain the total kinetic energy.

\(\begin{align}{K_T} &= K + {K_1}\\{K_T} &= \frac{1}{5}M{v^2} + \frac{1}{2}M{v^2}\end{align}\)

\({K_T} = \frac{7}{{10}}M{v^2}\)

Substitute the values in the above equation.

\(\begin{align}{K_T} &= \frac{7}{{10}} \times 7.25{\rm{ kg}} \times {\left( {3.10{\rm{ m/s}}} \right)^2}\\{K_T} &= 48.77{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ J}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\{K_T} &= 48.77{\rm{ J}}\end{align}\)

Thus, the total kinetic energy of a bowling ball is \({\rm{48}}{\rm{.77 J}}\).

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Most popular questions from this chapter

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

(a) so that it was also rotating?

(b) so that it wasn't rotating?

(c) It makes no difference.

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

(II) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, (a) that due to its daily rotation about its axis, and (b) that due to its yearly revolution about the Sun. (Assume the Earth is a uniform sphere with\({\bf{mass}} = {\bf{6}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{24}}}}{\bf{ kg}}\),\({\bf{radius}} = {\bf{6}}{\bf{.4 \times 1}}{{\bf{0}}^{\bf{6}}}{\bf{ m}}\)and is\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{\bf{8}}}{\bf{ km}}\)from the Sun.)

Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an axis through its center?

(a) The solid one.

(b) The hollow one.

(c) Both the same.

A uniform rod of mass M and length l can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as in Fig. 8–63. The rod is held horizontally and then released. At the moment of release, determine (a) the angular acceleration of the rod, and (b) the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown. [Hint: See Fig. 8–20g.]

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