Circular Motion

Question

A 7 kg mass moves in a circular path of 5 m radius on a frictionless horizontal table. It is attached to a string that passes through a frictionless hole in the center of the table. A second 87 kg mass is attached to the other end of the string. The acceleration of gravity is 9.8 m/s 2 .

a) Determine the period (the time for one revolution). Answer in units of s.

Answer 1

Let Mass 1 be the one on the table

Let Mass 2 be the one on the other end

Mass 1 Equation of Motion

T = M1a1

Mass 2 Equation of Motion

-T + m2g = m2a2

Note acceleration a1 = a2 = a (Same acceleration same variable, interconnected bodies via a single string)

Therefore

T = m1a Equation 1

-T + m2g = m2a Equation 2

Sub equation 1 onto equation 2

-(m1a) + m2g = m2a

a = m2g / m1+m2

Please note a1 = a = V^2/R (Centripetal acceleration of mass 1)

V^2 = (m2g/m1+m2)*R Equation 3

Now,

Circular motion Velocity

V = w * R

w = V/R Equation 4

Also,

w = θ/t

θ = 2π rad (Full revolution)

w = 2π / t

t = 2π/w Equation 5

Sub in Equation 4 into Equation 5

t = 2π/V/R = 2πR/V Equation 6

Take Equation 3 and solve for V

V^2 = (m2g/m1+m2)*R Equation 3

V^2 = (87*9.81/7+87)*5

V = 6.73 m/s

Now use Equation 6 to solve for the period

t = 2πR/V = 2π*5/6.73 = 4.66 seconds