A thin, uniform rod with negligible mass and length 0.230m is attached to the fl

Question

A thin, uniform rod with negligible mass and length 0.230m is attached to the floor by a frictionless hinge at point P (shown in the figure). A horizontal spring with force constant k = 4.80N/m connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B = 0.310T directed into the plane of the figure. There is current I = 6.40A in the rod, in the direction shown.

a) Calculate the magnitude of the torque due to the magnetic force on the rod, for an axis at P.

b) Calculate its direction (clockwise or counterclockwise)

c) Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? EXPLAIN ANSWER

d) When the rod is in equilibrium and makes an angle of 53.0 ? with the floor, is the spring stretched or compressed?

e) How much energy is stored in the spring when the rod is in equilibrium?

Explanation / Answer

a) T = F*(L/2) = I*L*B*(L/2) = 6.4*0.23*0.31*0.115 = 52.477*10^?3 Nm

b) clock wise

c) correct

d) stretched

e) torque T = k*x*L*sin53

x = T/(K*Lsin53) = 52.477*10^?3/(4.8*0.23*sin53) = 59.518*10^?3 m

U = 0.5*k*x^2 = 0.5*4.8*(59.518e?3^2) = 8.502*10^?3 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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