A mass of 3.0 kg is tied via a light string to the center of a table. A second m

Question

A mass of 3.0 kg is tied via a light string to the center of a table. A second mass of 5.0 kg is tied, also via a light string, to the first mass. Both masses are made to undergo uniform circular motion at a rate of 10 revolutions per second. If the first mass is 1.2 m from the center of the table and the second mass is 2.0 m from the center of the table, what is the tension in both strings?

A mass of 3.0 kg is tied via a light string to the center of a table. A second mass of 5.0 kg is tied, also via a light string, to the first / mass. Both masses are made to undergo uniform circular motion at a rate of 10 revolutions per second. If the first mass is 1.2 m from the center of the table and the second mass is 2.0 m from the center of the table, what is the tension in both strings? 6.

Explanation / Answer

T = mV^2/R

T1 = 3*(2*3.14*10)^2/1.2 ( 10 revolutions per second = 2pi*10 rad/s)

=> T1 = 9859.6 N

T2 = 5*(2*3.14*10)^2/2 = 9859.6 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.