Question

1. Extending the Particle in Uniform Circular Motion Model

A light string can support a station-ary hanging loadof$\mathbf{25}\mathbf{.}\mathbf{0}\mathit{k}\mathit{g}$beforebreaking. An object of mass

$\mathit{m}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{k}\mathit{g}$attached to the string rotates on a friction-less, horizontal table in a circle of radius

$\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathit{m}$, and the other end of the string is held fixed as in Figure P6.1. What range of speeds can the object have before the string breaks?

Short answer

As a result, the mass’s greatest speed is $8.08m/s$and its minimum speed is.$0m/s$

As a result, the mass tied to the string has a r$0\le v\le 8.08m/s$age of speed.

Step-by-step solution

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The string is capable of carrying the highest amount of weight. As a result, the string maximum tension is the weight of the suspended.

The centripetal force acting on the circulating object is equal to the tension in the string linked to the object. To compute the maximum speed of the object, multiply the maximum tension in the string by the centripetal force.

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Before breaking, the string would support.$25.0kg$ As a result, the string’s maximum tension is equal to the weight of the hanging object.

$T_{max}=Mg$

Here, M is the mass of the hanging weight, while denotes gravity.

Substitute$25.0kgforM9.8m/s^{2}forg$

$\begin{array}{l}{T}_{max}=(25.gkg)(9.8m/s^2)\\ =245N\end{array}$

The following equation describes the centripetal force acting on an object resolving with radius$r$ .

$F_e=\frac{m{v}^2}{r}$

Here, $v$is speed of the object.

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The tension in the string is balanced with the centripetal force exerted on the item because the tension in the string provides the centripetal acceleration.

$\begin{array}{l}\sum F_e=T\\ T=\frac{m{v}^2}{r}\end{array}$

Rearrange equation for.$v$

$v=\sqrt{\frac{rT}{m}}$

Substitute $0.800mforr,245NforTand3.0kgform$in equation and solve for $v$.

$\begin{array}{l}v=\sqrt{\frac{(0.800m)\left(245N\right)}{3.0kg}}\\ =8.08m/s\end{array}$

As a result, the mass’s greatest speed is and its minimum speed is.$0m/s$

As a result, the mass tied to the string has a $0\le v\le 8.08m/s$rage of speed.