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Expert-verified Found in: Page 132 ### Physics Principles with Applications

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.

See the step by step solution

## Step 1. Understanding the application of Newton’s second law in the bucket motion

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.

## Step 2. Identification of given data

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, $g=9.81\mathrm{m}/{\mathrm{s}}^{2}$.

## Step 3. Representation of the forces on the bucket at the bottom of the circle

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.

## Step 4. (a) Determination of the speed of the bucket at the bottom of the circle

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.

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

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

## Step 5. Representation of the forces on the bucket at the top of the circle

The forces on the bucket can be shown as, From the Newton’s second law, the equation can be expressed as,

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

## Step 6. (b) Determination of the speed of the bucket at the top of the circle

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.

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

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. ### Want to see more solutions like these? 