Question

A bucket of mass 2.00kg 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 25.0n. (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?

Answer 1

Answer:

(a). The speed of the bucket at the lowest point of its motion is $\boxed{1.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

(b). The speed of the bucket at the highest point to prevent the rope from being slack is $\boxed{3.43\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the bucket is $\boxed{1.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

The radius of the vertical circle is $\boxed{3.43\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

The tension in the rope at the lowest point is $25\,{\text{N}}$.

Concept:

Part (a):

The tension in the bucket at the lowest point of its motion is balanced by the weight and the centripetal force acting in the opposite direction of tension.

Write the expression for the forces acting on the bucket at the lowest point.

 $\begin{aligned}T&= {\text{weight}} + {\text{centripetal}}\,{\text{force}} \hfill \\T&= {\text{mg}} + \frac{{m{v^2}}}{r} \hfill\\\end{aligned}$

Substitute the values of the forces in above expression.

 $\begin{aligned}25&= \left( {2 \times 9.8} \right) + \frac{{2 \times {v^2}}}{{1.20}} \\ {v^2}&= \frac{{\left( {25 - 19.6} \right)1.20}}{2} \\ &= \sqrt {3.24} \,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&=1.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the speed of the bucket at the lowest point in the circle is $\boxed{1.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Part (b):

At the top point of the circle, the minimum velocity of the bucket should be such that the centripetal force becomes equal to the weight of the bucket.

The expression for the force balancing at top point is.

 $\begin{aligned}{\text{weight}}&= {\text{centripetal}}\,{\text{force}}\\{\text{mg}} &= \frac{{m{{\left( {v'} \right)}^2}}}{r}\\\end{aligned}$

Substitute the values in equation.

 $\begin{aligned}2 \times 9.8 &= \frac{{2 \times {{\left( {v'} \right)}^2}}}{{1.20}} \\ v'&= \sqrt {9.8 \times 1.20}\\ &=\sqrt {11.76} \,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.43\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the speed of the bucket at the highest point to prevent the rope from being slack is $\boxed{3.43\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

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Answer Details:

Grade:Senior School

Chapter:Vertical circular motion

Subject:Physics

Keywords: Bucket, lowest point, circular motion, vertical, speed of bucket, top of circle, tension, centripetal force, weight, forces.

Answer 2

(a) The speed of the bucket is 1.8 m/s.

(b) The speed of bucket to undergo the non-slacking condition is 3.43 m/s

Given data:

The mass of bucket is, m = 2 kg.

The radius of vertical circle is, r = 1.20 m.

The magnitude of tension is, T = 25 N.

(a)

Since the bucket is attached with a rope, providing it a whirling motion. Then three forces are acting on the system. They are,

1. Tension force by the rope
2. Centripetal force on bucket due to whirling motion in a vertical   circle.
3. Weight of bucket.

At the lowest point of vertical motion of bucket, the tension on the bucket at lowest is balanced by the weight and the centripetal force acting on the bucket. Therefore,

$T = W + F_{c}\\\\T = mg + \dfrac{mv^{2}}{r}\\25 = 2 \times 9.8+ \dfrac{2 \times v^{2}}{1.20}\\\\25-19.6 = \dfrac{2 \times v^{2}}{1.20} \\v = 1.8 \;\rm m/s$

Thus, the speed of the bucket is 1.8 m/s.

(b)

The slacking rope occurs when the centripetal force acting on the bucket acts in a direction opposite to weight of bucket. Therefore for non- slacking condition, the weight is balanced by the centripetal force. Then,

$W = F_{c}\\mg = \dfrac{mv'^{2}}{r}\\\\g = \dfrac{v'^{2}}{r}\\\\9.8 = \dfrac{v'^{2}}{1.20}\\v' = 3.43 \;\rm m/s$

Thus, the speed of bucket to undergo the non-slacking condition is 3.43 m/s

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