Physics, 28.07.2019 07:30 juliaduenkelsbu

Abucket 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?

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Answer from: mal5546

(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 $\boxed{3.43\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Learn More:

An object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second link A 1200 kg steel beam is supported by the two ropes calculate the tension in the rope link A car traveling at speed v takes distance d to stop after the brakes are appliedhttps://link

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 from: Quest

The answer is b.the gaseous state is electrically positive, while the plasma state is electrically negative.

Answer from: Quest

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

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