Respuesta :
Answer:
torque is 0.2324 N-m
Explanation:
given data
mass m1 = 1 kg
mass m2 = 2 kg
length r2 = 1.0 m
angular velocity = 20 rpm = 20 × 2π/60sec = 2.092 rad/s
to find out
What torque will bring the balls to a halt in 5.0 s
solution
we know center of mass of rod is = m1r1 + m2r2 / (m1+m2)
= 1 (0) + 2(1.0) / (1+2)
= 2/3 = 0.666 m
so distance from center of rod (d) = 1 - 0.66 = 0.333
so that we say
distance of axis rotation to m1 is r1 i.e = 0.666
distance of axis rotation to m2 is r2 i.e = 0.334
so we now find moment of inertia of rod about center of mass
inertia = m1r1² + m2r2²
inertia = 1 (0.666)² + 2 ( 0.334)²
inertia = 0.4435 + 0.1115 = 0.555 kg m²
so
angular acceleration = angular velocity / time
angular acceleration = 2.0932 / 5 = 0.4187 rad/s²
we apply torque formula
torque = inertia ×angular acceleration
torque = 0.555 × 0.4187
torque is 0.2324 N-m
The torque that will bring the ball to a halt is 0.233 N.m.
The given parameters;
- mass of the balls, m = 1 kg, 2 kg
- length, r = 1 m
- angular speed, = 20 rpm
- time of motion, t = 5 s
The distance of each mass from the center mass is calculated as;
[tex]X_c_m = \frac{1(0) + 2(1)}{1 + 2} = 0.666 \ m[/tex]
the position of the first mass from center = 0.666 m
the position of the second mass = 1 - 0.666 m = 0.334 m
The moment of inertia of the balls is calculated as follows;
[tex]I = m_1 r_1^2 + m_2r^2\\\\I = 1(0.666)^2 + 2(0.334)^2\\\\I = 0.555 \ kgm^2[/tex]
The angular acceleration of the balls is calculated as follows;
[tex]a_c = \frac{\omega }{t} \\\\a_c = 20 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{rev} \times \frac{1\min}{60 \ s} \times \frac{1}{5 \ s} = 0.42 \ rad/s^2[/tex]
The torque that will bring the ball to a halt is calculated as follows;
[tex]\tau \ = I \alpha \\\\\tau = 0.555 \times 0.42 \\\\\tau = 0.233 \ Nm[/tex]
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