When the mass is at the top the force equation is mω^{2}R = mg +T where T is the tension in the cord. At the bottom the force equation is T - mg = mω^{2}R. To find the period of motion solve for ω which is 2π/Period. Take the two equations and solve for T by adding the equations. This gives T= mω^{2}R = 1.7• (2π10)^{2}(.7) = 476π^{2} = 4697.9 N= 4.697 kN. The period of rotation is 1/10 or .1 sec the rotational velocity is v=Rω =.7(20 π) so the rotational velocity is 43.98 m/s. One rotation is 2π rad Θ=ωt so the time to complete 75 rotations = 75Θ/ω which is 75• 2π/2π10 and t= 75/10 = 7.5s

Mimi ..

asked • 7d# plzz helppp urgenttttt

A 1.7kg ball is spun around in circular motion such that its radius is 0.7m and frequency is 10Hz.

a. What is the period of its rotation?

b. How much time will be required to complete 75 rotations?

c. What is the rotational velocity of the ball?

d. What is the Tension Force in the string?

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Mimi ..

thankk you6d