Question 17.SP.3: A sphere, a cylinder, and a hoop, each having the same mass ......

STRATEGY: You are given two positions, want to find the velocities, and the friction force F in rolling motion does no work, so use the conservation of energy. First solve the problem in general terms, and then find the results for each body. Denote the mass by m, the centroidal moment of inertia by \bar{I} and the radius by r.

MODELING: Choose the rolling object as your system and model it as a rigid body. Because each body rolls, the instantaneous center of rotation is located at C (Fig. 1). Free-body diagrams of the system at the two locations are shown in Fig. 2.

ANALYSIS:

Conservation of Energy.

T_1+V_{g_1}+V_{e_1}=T_2+V_{g_2}+V_{e_2}        (1)

Potential Energy.

Because there is no spring in the system, V_{e_1}=V_{e_2}=0. If you place your datum at the center of mass of the system when it is at position 2, you have V_{g_2} = 0 and V_{g_1} = mgh.

Kinetic Energy.

T_1=0

T_2=\frac{1}{2} m \bar{v}^2+\frac{1}{2} \bar{I} \omega^2

Kinematics. You need to relate \bar{v} and ω using kinematics. Because each body rolls, the instantaneous center of rotation is located at C (Fig. 1), which gives

\omega=\frac{\bar{v}}{r}

Substituting this into T_2 gives

T_2=\frac{1}{2} m \bar{v}^2+\frac{1}{2} \bar{I}\left(\frac{\bar{v}}{2}\right)^2=\frac{1}{2}\left(m+\frac{\bar{I}}{r^2}\right) \bar{v}^2

Substituting these energy expressions into Eq. (1) gives

0 + mgh + 0 = \frac{1}{2}\left(m+\frac{\bar{I}}{r^2}\right) \bar{v}^2 + 0 + 0

Solving for the speed at position 2, you find

\bar{v}^2=\frac{2 g h}{1+\bar{I} / m r^2}

Velocities of Sphere, Cylinder, and Hoop. Introducing the particular expressions for \bar{I}, you obtain

Sphere:                     \bar{I}=\frac{2}{5} m r^2                     \bar{v}=0.845 \sqrt{2 g h} ◂

Cylinder:                     \bar{I}=\frac{1}{2} m r^2                    \bar{v}=0.816 \sqrt{2 g h} ◂

Hoop:                     \bar{I}= m r^2                    \bar{v}=0.707 \sqrt{2 g h} ◂

REFLECT and THINK: Comparing the results, we note that the velocity of the body is independent of both its mass and radius. However, the velocity does depend upon the quotient of \bar{I} / m r^2=\bar{k}^2 / r^2, which measures the ratio of the rotational kinetic energy to the translational kinetic energy. Thus, the hoop, which has the largest \bar{k} for a given radius r, attains the smallest velocity.

Let us compare the results with the velocity attained by a frictionless block sliding through the same distance. The solution is identical to the previous solution except that ω = 0; we find \bar{v}=\sqrt{2 g h}. So, all the rolling objects are slower than one moving down a frictionless surface.