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12CQ

Expert-verifiedFound in: Page 324

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**If a small sphere of mass M were placed at the end of the rod in Figure 10.21, would the result for be greater than, less than, or equal to the value obtained in Example 10.11?**

The angular speed of rod with sphere at its end is less than the angular speed of the rod.

**the moment of inertial plays the same role in rotational motion, as mass plays in linear motion. For a system, the moment of inertia is defined as the product of the total mass of the system and the sum of square of distance of different particles, constituting the system, from the axis of rotation.**

The moment of inertia of the rod is given as:

${I}_{r}=\frac{M{L}^{2}}{3}$

Here, M is the mass of rod and L is the length of rod.

Apply the work-energy theorem between horizontal position and lowest position of rod to find the angular speed of rod.

${K}_{f}-{K}_{i}={W}_{r}\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\left(2\right)$

Here, K is the initial rotational kinetic energy of rod and its value is zero and K_{f} is the initial rotational kinetic energy of rod.

The work done by weight of rod for distance $\frac{L}{2}$ is ${W}_{L/2}=Mg\left(\frac{L}{2}\right)$

For the above value of work, equation (2) becomes-

$\frac{1}{2}{I}_{r}{\omega}_{r}^{2}-0=Mg\left(\frac{L}{2}\right)\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(\frac{M{L}^{2}}{3}\right){\omega}_{r}^{2}=Mg\left(\frac{L}{2}\right)\phantom{\rule{0ex}{0ex}}{\omega}_{r}=1.73\sqrt{\frac{g}{L}}\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\xb7\left(1\right)$

The moment of inertia of the combined sphere and rod is given as:

${I}_{t}={I}_{r}+M{L}^{2}\phantom{\rule{0ex}{0ex}}=\frac{M{L}^{2}}{3}+M{L}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4M{L}^{2}}{3}$

Here, M is the mass of rod and L is the length of rod.

Apply the work-energy theorem between horizontal position and lowest position of rod to find the angular speed of rod.

${K}_{f}-{K}_{i}={W}_{t}$

Here, K_{i} is the initial rotational kinetic energy of rod with sphere and its value is zero. The work done by weight of rod for distance $\frac{L}{2}$ is ${W}_{L/2}=Mg\left(\frac{L}{2}\right)$ and work done by the weight of sphere for distance L is mgL .

$\frac{1}{2}{I}_{t}{\omega}_{t}^{2}-0=Mg\left(\frac{L}{2}\right)+MgL\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(\frac{4M{L}^{2}}{3}\right){\omega}_{t}^{2}=Mg\left(\frac{3L}{2}\right)\phantom{\rule{0ex}{0ex}}{\omega}_{t}=1.06\sqrt{\frac{g}{L}}............................................\left(3\right)$

The equation (1) and equation (3) clearly shows that angular speed of rod is more than the angular speed of system of rod and sphere. The angular speed of rod with sphere, at its end, will be less than the angular speed of rod.

Therefore, the angular speed of rod with sphere at its end is less than the angular speed of the rod.

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