A small object of mass m moves in a horizontal circle of radius r on a rough tab

Question

A small object of mass m moves in a horizontal circle of radius r on a rough table. It is attached to a horizontal string fixed at the center of the circle. The speed of the object is initially v0. After completing one full trip around the circle, the speed of the object is 0.5v0.
(a) Find the energy dissipated by friction during that one revolution in terms of m, v0, and r. (Use any variable or symbol stated above along with the following as necessary: g.)
?Etherm =

(b) What is the coefficient of kinetic friction? (Use any variable or symbol stated above along with the following as necessary: g.)
µk =

(c) How many more revolutions will the object make before coming to rest?

Explanation / Answer

A) Final Energy - Initial Energy = -Work by Non-conservativeforces The initial energy is: Ke = 1/2mVo2 The final energy is: Kef = 1/2*m*(1/2*Vo)2 =1/8*m*Vo2 Therefore, the energy dissipated by friction is: -(1/8mV02 -1/2mVo2) = 3/8mVo2 B) We know the work done by frictionis: W =2r**mg         where 2r isthe circumerference of the the circle 3/8mVo2 =2r*mg = 3Vo2 /(16rg) C) Final Energy - Initial Energy = -Workby Non-conservative forces 0 - 1/8mV02=-(3Vo2 / (16rg) )*mg*x x = 2/3*r (2/3)*r / (2r) =1/3 of acircle A) Final Energy - Initial Energy = -Work by Non-conservativeforces The initial energy is: Ke = 1/2mVo2 The final energy is: Kef = 1/2*m*(1/2*Vo)2 =1/8*m*Vo2 Therefore, the energy dissipated by friction is: -(1/8mV02 -1/2mVo2) = 3/8mVo2 B) We know the work done by frictionis: W =2r**mg         where 2r isthe circumerference of the the circle 3/8mVo2 =2r*mg = 3Vo2 /(16rg) C) Final Energy - Initial Energy = -Workby Non-conservative forces 0 - 1/8mV02=-(3Vo2 / (16rg) )*mg*x x = 2/3*r (2/3)*r / (2r) =1/3 of acircle (2/3)*r / (2r) =1/3 of acircle

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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