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An air puck of mass 0.029 kg is tied to
a string and allowed to revolve in a circle
of radius 1.7 m on a frictionless horizontal
surface. The other end of the string passes
through a hole in the center of the surface,
and a mass of 1.5 kg is tied to it, as shown.
The suspended mass remains in equilibrium
while the puck revolves on the surface.
0.029 kg
1.7 m
1.5 kg
What is the magnitude of the force that
maintains circular motion acting on the puck?
The acceleration due to gravity is 9.81 m/s?.
Answer in units of N.
part 2 of 2
What is the linear speed of the puck?
Answer in units of m/s.

An Air Puck Of Mass 0029 Kg Is Tied To A String And Allowed To Revolve In A Circle Of Radius 17 M On A Frictionless Horizontal Surface The Other End Of The Stri class=

Sagot :

onyebuchinnaji

(1) The magnitude of the force that maintains circular motion acting on the puck is 14.7 N.

(2) The linear speed of the puck is 29.35 m/s.

The given parameters:

  • Mass of air puck, m₁ = 0.029 kg
  • Suspended mass, m₂ = 1.5 kg
  • Radius of the circle, r = 1.7 m

The magnitude of the force that maintains circular motion acting on the puck;

[tex]F = \frac{m_1 v^2}{r}[/tex]

The tension on the string due to the suspended masse is calculated as follows;

[tex]T = m_2 g[/tex]

Since the two masses are in equilibrium;

[tex]\frac{m_1 v^2}{r} = m_2g\\\\m_1v^2 = m_2g r\\\\v^2 = \frac{m_2g r}{m_1} \\\\v= \sqrt{\frac{m_2g r}{m_1} } \\\\v = \sqrt{\frac{1.5 \times 9.8 \times 1.7}{0.029} } \\\\v = 29. 35 \ m/s[/tex]

The magnitude of the force that maintains circular motion acting on the puck;

[tex]F = \frac{m_1v^2}{r} \\\\F = \frac{0.029 \times (29.35)^2}{1.7} \\\\F = 14.7 \ N[/tex]

Learn more about centripetal force here: https://brainly.com/question/20905151

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