The diagram shows a ring of radius 2.0m and mass 5.0kg being subjected to a torq

Question

The diagram shows a ring of radius 2.0m and mass 5.0kg being subjected to a torque tau rightarrow of 30.0 N-m pointing into the paper. Find the direction and magnitude of the resulting angular acceleration. The diagram shows a rod spinning faster and faster in such a direction that the end marked A is travelling into the paper and the other end marked B is out of the paper. Draw in the vector torque tau rightarrow. The ring of Problem 1 carries a 3.0 A current in the clockwise direction, and it is in a magnetic field of 0.5T pointing upward as shown. Find (a) the magnitude and direction of the magnetic moment vector mu rightarrow, (b)the magnitude and direction of the torque vector tau rightarrow, (c) the angular acceleration of the ring and (d) the direction of acceleration of the point A at the top of the ring.

Explanation / Answer

1) given

r = 2m

m = 5 kg

T = 30 N.m

we know, T = I*alfa

==> alfa = T/I

= T/(m*R^2)

= 30/(5*2^2)

= 1.5 rad/s^2

direction : into the page

2) draw an arrow downward

3)

a) mue = I*A

= 3*pi*2^2

= 75.4 A.m^2

direction : into the page

b) T = mue*B*sin(theta)

= 75.4*0.5*sin(90)

= 37.7 N.m

direction : rightward

c) moment of inertia about perpendicualr axis, I = 0.5*m*R^2

= 0.5*5*2^2

= 10 kg.m^2

so, alfa = T/I

= 37.7/10

= 3.77 rad/s^2

d) rightward (along +x axis)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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