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Q10.

Expert-verifiedFound in: Page 132

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.20 m. At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N.**

**(a) Find the speed of the bucket.**

**(b) How fast must the bucket move at the top of the circle so that the rope does not go slack?**

(a) The speed of the bucket at the bottom of the circle is $1.797\mathrm{m}/\mathrm{s}$.

(b) The speed of the bucket at the top of the circle is $3.43\mathrm{m}/\mathrm{s}$. The bucket will move faster when the rope does not slack.

**A bucket is tied with a rope, and it executes circular motion.**** From Newton’s second law, the net force acting in the vertical direction will be equated to the force acting on the bucket. ****This force is in the force of centripetal force. **

**By using these equations, the speed at the top or bottom of the circle can be evaluated.**

The given data can be listed below as,

- The radius of circle is, $r=1.20\mathrm{m}$.
- The mass of the bucket is, $m=2\mathrm{kg}$.
- The tension in the rope is, $T=25\mathrm{N}$.
- The acceleration due to gravity is, $g=9.81\mathrm{m}/{\mathrm{s}}^{2}$.

** **

The forces on the bucket can be shown as,

Here, *mg *is the weight of the bucket, ${F}_{\mathrm{T}}$ is the tension force in the rope,* m* is the mass of the bucket.

** **

From the Newton’s second law, the equation can be expressed as,

$\begin{array}{c}\sum {F}_{y}={F}_{\mathrm{c}}\\ {F}_{\mathrm{T}}-mg=m{a}_{\mathrm{c}}\\ \frac{m{v}^{2}}{r}={F}_{\mathrm{T}}-mg\\ v=\sqrt{\frac{r\left({F}_{\mathrm{T}}-mg\right)}{m}}\end{array}$

Here, *v* is the bucket’s speed, ${F}_{\mathrm{c}}$ is the centripetal force, ${a}_{\mathrm{c}}$ is the centripetal acceleration of the bucket.

Substitute the values in the above equation.

$\begin{array}{c}v=\sqrt{\frac{1.20\mathrm{m}\left(25\mathrm{N}\left(\frac{1\mathrm{kg}\xb7\mathrm{m}/{\mathrm{s}}^{2}}{1\mathrm{N}}\right)-2\mathrm{kg}\times 9.81\mathrm{m}/{\mathrm{s}}^{2}\right)}{2\mathrm{kg}}}\\ v=\sqrt{3.228}\mathrm{m}/\mathrm{s}\\ v=1.797\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the bucket at the bottom of the circle is $1.797\mathrm{m}/\mathrm{s}$.

The forces on the bucket can be shown as,

From the Newton’s second law, the equation can be expressed as,

$\begin{array}{c}\sum {F}_{y}={F}_{\mathrm{c}}\\ {F}_{\mathrm{T}}+mg=m{a}_{\mathrm{c}}\\ \frac{m{v}^{2}}{r}={F}_{\mathrm{T}}+mg\\ v=\sqrt{\frac{r\left({F}_{\mathrm{T}}+mg\right)}{m}}\end{array}$

** **

The rope will not go to slack. So, the tension in the rope will become zero.

Now, the above equation can be written as,

$\begin{array}{c}v=\sqrt{\frac{r\left(0+mg\right)}{m}}\\ =\sqrt{\frac{rmg}{m}}\\ =\sqrt{rg}\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}v=\sqrt{1.20\mathrm{m}\times \left(9.81\mathrm{m}/{\mathrm{s}}^{2}\right)}\\ v=\sqrt{11.772}\mathrm{m}/\mathrm{s}\\ v=3.43\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the bucket at the top of the circle is $3.43\mathrm{m}/\mathrm{s}$.

From the above two cases, (a) and (b), it can be concluded that the bucket will move faster at the top of the circle when the rope does not slack.

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