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Q68P

Expert-verifiedFound in: Page 291

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Two uniform solid spheres have the same mass of${\mathbf{\text{1.65 kg}}}$ ****, but one has a radius of ${\mathbf{\text{0.226 m}}}$**** and the other has a radius of ****. Each can rotate about an axis through its center. (a) What is the magnitude **** of the torque required to bring the smaller sphere from rest to an angular speed of ${\mathbf{\text{317 rad/s}}}$**** in ${\mathbf{\text{15.5 s}}}$****? (b) What is the magnitude ****of the force that must be applied tangentially at the sphere’s equator to give that torque? What are the corresponding values of (c) **** $\tau $and (d) ****for the larger sphere?**

- The magnitude of the torque required to bring the smaller sphere from rest to an angular speed of $317\text{rad/s}$in $15.5\text{s}$ is $0.689\text{N}\xb7\text{m}$.
- The magnitude of the force that must be applied tangentially at the sphere’s equator to give the torque is .$3.05\text{N}$
- The corresponding value for the larger sphere is $9.84\text{N}\xb7\text{m}$
- The corresponding value for the larger sphere is $11.52\text{N}$

Mass of the solid spheres, $\text{m}=1.65\text{kg}$

The radius of the smaller sphere, $\text{r}=0.226\text{m}$

The radius of the larger sphere, $\text{R}=0.854\text{m}$

Initial angular speed, ${\omega}_{0}=0\text{rad/s}$

Final angular speed, $\omega =317\text{rad/s}$

Time of rotation, $\text{t}=15.5\text{s}$

**A body rotating with a net angular acceleration will always have net torque which is a force applied on the body about the axis of rotation, acting on it. Thus, for a body in rotation, the force due to this torque acting on any tangential point of the body has the same magnitude value as the given torque. The torque applied at the center is thus given by the force magnitude product with the distance between the axis and the point at which the force acts.**

Formulae:

The torque acting on a body due to the rotational analog of Newton’s second law,

$\tau =I\alpha $ (1)

where is the moment of inertia of the body passing through its central axis and is the angular acceleration of the body.

The moment of inertia of the sphere about its central axis, $I=\frac{2}{5}m{r}^{2}$ (2)

The angular acceleration of the body in rotational motion, $\alpha =\frac{\omega -{\omega}_{0}}{t}$ (3)

The force acting on a body due to its torque, $F=\frac{\tau}{r}$ (4)

Using the given data in equation (3), the angular acceleration of the spheres can be given as:

$\begin{array}{rcl}\alpha & =& \frac{317\text{rad/s}-0\text{rad/s}}{15.5\text{s}}\\ & =& 20.45{\text{rad/s}}^{2}\\ & & \end{array}$

Now, the magnitude of the torque acting on the smaller sphere can be given using the above value with equation (2) and the given data in equation (1) as follows:

$\tau =\left(\frac{2}{5}m{r}^{2}\right)\alpha \phantom{\rule{0ex}{0ex}}=\frac{2}{5}\left(1.65\text{kg}\right){\left(0.226\text{m}\right)}^{2}\left(20.45\text{rad/s}\right)\phantom{\rule{0ex}{0ex}}=0.689\text{N}\xb7\text{m}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the torque is .$0.689\text{N}\xb7\text{m}$

Now, using the above torque value in equation (4), the magnitude of the force that must be applied tangentially at the smaller sphere’s equator to give the torque can be given as follows:

$\begin{array}{rcl}F& =& \frac{0.689\text{N}\xb7\text{m}}{0.226\text{m}}\\ & =& 3.05\text{N}\\ & & \end{array}$

Hence, the value of the force is $3.05\text{N}$

Now, the magnitude of the torque acting on the larger sphere can be given using the above value with equation (2) and the given data in equation (1) as follows:

$\begin{array}{rcl}\tau \text{'}& =& \left(\frac{2}{5}m{R}^{2}\right)\alpha \\ & =& \frac{2}{5}\left(1.65\text{kg}\right){\left(0.854\text{m}\right)}^{2}\left(20.45\text{rad/s}\right)\\ & =& 9.84\text{N}\xb7\text{m}\\ & & \end{array}$

Hence, the value of the torque is .$9.84\text{N}\xb7\text{m}$

Now, using the above torque value in equation (4), the magnitude of the force that must be applied tangentially at the larger sphere’s equator to give the torque can be given as follows:

$F\text{'}=\frac{9.84\text{N}\xb7\text{m}}{0.854\text{m}}\phantom{\rule{0ex}{0ex}}=11.52\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the force is $11.52\text{N}$

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