Physics College

A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant speed of 6 m/s. a) Calculate the magnitude of the normal force acting on the car when it is on the top of the circle and when it is on the bottom of the circle, respectively. b) What is the minimum speed the car needs to go around the circle without falling off

Answers

Answer 1

Answer:

a)[tex]|N|=9.83\: N[/tex]  at the top

[tex]|N|=33.37\: N[/tex] at the botton

b) The minimum velocity will be [tex]v=4.43\: m/s[/tex].

Step-by-step explanation:

a) Using the second Newton's law, at the top of the circle we have.

[tex]\Sigma F=ma_{c}[/tex]

The forces at the top are the weight and the normal force.

[tex]W-N=m\frac{v^{2}}{R}[/tex]

[tex]mg-N=m\frac{v^{2}}{R}[/tex]

[tex]N=mg-(m\frac{v^{2}}{R})[/tex]

[tex]N=1.2*9.81-(1.2\frac{6^{2}}{2})[/tex]        

[tex]N=-9.83\: N[/tex]  

[tex]|N|=9.83\: N[/tex]      

At the botton of the circle we have:

[tex]N-W=m\frac{v^{2}}{R}[/tex]

[tex]N-mg=m\frac{v^{2}}{R}[/tex]

[tex]N=1.2*9.81+(1.2\frac{6^{2}}{2})[/tex]  

[tex]|N|=33.37\: N[/tex]

b) If we do the normal force equal to zero we can find the minimum velocity, which means:

[tex]W-0=m\frac{v^{2}}{R}[/tex]

[tex]mg=m\frac{v^{2}}{R}[/tex]

[tex]v=\sqrt{Rg}[/tex]

[tex]v=\sqrt{2*9.81}[/tex]

Therefore, the minimum velocity will be [tex]v=4.43\: m/s[/tex].

I hope it helps you!

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