Question

The figure shows a rigid structure consisting of a circular hoop of radius R and mass m1 and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of s. If R = m and m = kg, calculate the angular momentum about that axis. At the instant of the figure, a kg particle P has a position vector r rightarrow of magnitude m and angle Q1 = degree and a velocity vector v rightarrow of magnitude m/s and angle theta 2 = degree, Force F rightarrow, of magnitude N and angle theta 3 = degree acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?

Answer 1

Part-A : given that,   mass of circular hoop, M = 3.8 kg

radius of circular hoop, R = 0.1 m

Time period of rotation, T = 2.6 sec

angular momentum about that axis is given as :

L = I_net                                                     { eq. 1}

where, I_net = total rotational inertia which is given by below :

Rotational inertia of hoop :

I_H = M R²                                               { eq. 2 }

Rotational inertia of just one radial rod (about the end touching the center of the hoop) :

I_1rod = (1 / 3) m R²                                    { eq. 3 }

Total rotational inertia :

I_net = I_H + 4 I_{1 rod}                                     { eq. 4 }

I_net = M R² + (4/3) m R²                                                     (from eq.2 & 3)

Because one rod is equal in mass to hoop :

I_net = (7/3) M R²                                                        { eq. 5 }

using eq. 1,

L = I_net

where, = angular velocity in terms of peiod = 2 / T

L = (7/3) M R² x (2 / T) = 14 M R² / 3 T

inserting the values in above eq.

L = 14 (3.14) (3.8 kg) (0.1 m)² / 3 (2.6 sec) = 1.67048 / 7.8 kg m² /s

L = 0.214 kg m² /s

Part-B : given that,

mass of a particle, m =3.2 kg

radius, r = 5.4 m                            at ₁ = 43 degree

velocity, v = 8.2 m/s                        at ₂ = 28 degree

force, F = 2.2 N                             at ₃ = 28 degree

All three vectors lie in the same plane. torque acting on the particle is given as ::

Torque, T = r x F                        { eq. 6 }

r. cos = (5.4 m) .cos 43 = 3.949

r. sin = (5.4 m) sin 43 = 3.682

F. cos = (2.2 N) cos 28 = 1.942

F. sin = (2.2 N) sin 28 = 1.032

T_x = 3.949 x 1.942 = 7.668

T_y = 3.682 x 1.032 = 3.779

T = ? T_x + T_y = ? 58.79 + 14.28 = 21.94 N-m