Question48. (II) A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.10 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?

Studdy Solution

STEP 1

1. The bucket has a mass of 2.00 kg.
2. The radius of the vertical circle is 1.10 m.
3. At the lowest point, the tension in the rope is 25.0 N.
4. We are using the acceleration due to gravity as $g = 9.81 \, \text{m/s}^2$ .
5. We need to find the speed of the bucket at the lowest point and the minimum speed at the top of the circle to prevent the rope from going slack.

STEP 2

1. Analyze forces at the lowest point of the circle.
2. Calculate the speed of the bucket at the lowest point.
3. Analyze forces at the top of the circle.
4. Calculate the minimum speed at the top of the circle.

STEP 3

Analyze forces at the lowest point of the circle.
At the lowest point, the forces acting on the bucket are the gravitational force and the tension in the rope. The net force provides the centripetal force required for circular motion.

STEP 4

Calculate the speed of the bucket at the lowest point.
The net force at the lowest point is given by:
$F_{\text{net}} = T - mg = \frac{mv^2}{r}$
Where: - $T = 25.0 \, \text{N}$ is the tension, - $m = 2.00 \, \text{kg}$ is the mass, - $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity, - $r = 1.10 \, \text{m}$ is the radius, - $v$ is the speed at the lowest point.
Rearrange to solve for $v$ :
$v^2 = \frac{r(T - mg)}{m}$
$v = \sqrt{\frac{r(T - mg)}{m}}$
Substitute the known values:
$v = \sqrt{\frac{1.10 \, (25.0 - 2.00 \times 9.81)}{2.00}}$
$v = \sqrt{\frac{1.10 \, (25.0 - 19.62)}{2.00}}$
$v = \sqrt{\frac{1.10 \times 5.38}{2.00}}$
$v = \sqrt{\frac{5.918}{2.00}}$
$v = \sqrt{2.959}$
$v \approx 1.72 \, \text{m/s}$

STEP 5

Analyze forces at the top of the circle.
At the top of the circle, the gravitational force and the tension both act downwards. To prevent the rope from going slack, the tension must be zero or greater. The gravitational force alone must provide the centripetal force.

STEP 6

Calculate the minimum speed at the top of the circle.
The condition for the rope not going slack is:
$mg = \frac{mv^2}{r}$
Canceling $m$ from both sides gives:
$g = \frac{v^2}{r}$
Rearrange to solve for $v$ :
$v^2 = gr$
$v = \sqrt{gr}$
Substitute the known values:
$v = \sqrt{9.81 \times 1.10}$
$v = \sqrt{10.791}$
$v \approx 3.28 \, \text{m/s}$
The speed of the bucket at the lowest point is approximately $1.72 \, \text{m/s}$ , and the minimum speed at the top of the circle to prevent the rope from going slack is approximately $3.28 \, \text{m/s}$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms