1) 1.5 rad / s 2 2) clockwise downward 3) ( a )37.7 A * m^ 2 into p aper ( b )18

Question

1) 1.5rad / s2  

2) clockwise downward

3) (a)37.7A*m^2   into paper   (b)18.9N!m   totheright   (c)1.89rad/s2   (d)out of the paper

1. The diagram shows a ring of radius 2.0m and mass 5.0kg being subjected to a torque tof 30.0 N-m pointing into the paper. Find the direction and magnitude of the resulting angular acceleration. 2. 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 Bis out of the paper. Drawin the vector torque t. 3. The ring of Problem l 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 pa, (b)the magnitude and direction of the torque vector t (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) I = m*r^2 = 5*2^2 = 20 kg.m^2

we know, T = I*alfa

==> alfa = T/I = 30/20 = 1.5 rad/s^2

2) dowward

3)
a) magnetic momen, M = I*A = 3*pi*2^2 = 37.7 A.m^2

direction is in to the paper.

b) T = M*B*sin(90)

= 37.7*0.5*1

= 18.85 N.m

direction is towards +x axis

c) I = 0.5*m*r^2 = 0.5*5*2^2 = 10 kg.m^2

T = I*alfa

==> alfa = T/I = 18.85/10 = 1.885 rad/s^2

d) out of the page.

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