The tension in the string holding the [tex]200\,{\text{g}}[/tex] puck is [tex]\boxed{360\,{\text{N}}}[/tex] .
Further Explanation:
Given:
The mass of the puck is [tex]200\,{\text{g}}[/tex] or [tex]0.2\,{\text{kg}}[/tex] .
The length of the string or the radius of the circular path is [tex]50.0\,{\text{cm}}[/tex] or [tex]0.5\,{\text{m}}[/tex] .
The angular momentum of the rotating puck is [tex]3.00\,{{{\text{kg}} \cdot {{\text{m}}^{\text{2}}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {{\text{m}}^{\text{2}}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex].
Concept:
As the puck revolves in a circle on the frictionless table, the puck experiences an outward centripetal force and this force is equal to the tension developed in the string holding the puck.
The angular momentum of the puck can be expressed as:
[tex]L = mvr[/tex]
Substitute the value of [tex]L[/tex] , [tex]m[/tex] and [tex]r[/tex] in above expression.
[tex]\begin{aligned}3 &= 0.2 \times v \times 0.5 \hfill\\v &= 30\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \hfill\\\end{aligned}[/tex]
Now, the centripetal force acting on the puck is balanced by the tension developed in the string.
Use the force balancing equation for the puck.
[tex]T = \dfrac{{m{v^2}}}{r}[/tex]
Substitute the values of [tex]m[/tex] , [tex]v[/tex] and [tex]r[/tex] in above expression.
[tex]\begin{aligned}T&= \frac{{0.2 \times {{\left( {30} \right)}^2}}}{{0.5}}\\&= \frac{{180}}{{0.5}}\\&= 360\,{\text{N}}\\\end{aligned}[/tex]
Thus, the tension in the string holding the [tex]200\,{\text{g}}[/tex] puck is [tex]\boxed{360\,{\text{N}}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Angular Momentum
Keywords: 200g puck, frictionless table, 50 cm long string, angular momentum, tension, L=mvr, centripetal force, mv^2/r, revolves in a circle, tension developed.