Consider a 8·kg rod with a length of 0.6·m and rotating about an axis passing through its center.
(a) What is its rotational inertia?
(B) Suppose the rod is rotating counterclockwise and speeding up with a rotational acceleration of 7·rad/s2. What is the net torque acting on the rod?
(a) I = (1/12) M L^2 = 0.24 kg*m^2
(b) I * 7 = 1.68 N-m
torque= I*angular acceleration
solve for torque. Now sign. It is very arbritrary, but most write clockwise as positive, counterclockwise as negative, for torques, and rotation. Follow what your teacher gives as law on this convention of signs.
l*angular accelertion
thank you! From here how would I get this:
Suppose the torque you found for the last part is being produced by a force applied at a point 0.15·m from the axis. If the direction of the force is perpendicular to a line from the axis to that point, what is the magnitude of the force?
Sorry I missed this day in class and this is not a strong subject for me!
Force * 0.15 m = 1.68 N*m
Solve for force
thank you so much!!!
To find the rotational inertia of the rod (a), we need to use the formula for the rotational inertia of a uniform rod:
I = (1/12) * m * L^2
where I is the rotational inertia, m is the mass of the rod, and L is the length of the rod.
Given:
mass (m) = 8 kg
length (L) = 0.6 m
Plugging in the values into the formula, we can calculate the rotational inertia:
I = (1/12) * (8 kg) * (0.6 m)^2
Simplifying the equation, we get:
I = (1/12) * 8 kg * 0.36 m^2
I = 0.24 kg * m^2
Therefore, the rotational inertia of the rod is 0.24 kg * m^2.
To find the net torque acting on the rod (b), we need to use the formula:
Torque (τ) = I * α
where τ is the torque, I is the rotational inertia, and α is the rotational acceleration.
Given:
rotational acceleration (α) = 7 rad/s^2
rotational inertia (I) = 0.24 kg * m^2 (from part a)
Plugging in the values into the formula, we can calculate the net torque:
τ = (0.24 kg * m^2) * (7 rad/s^2)
Simplifying the equation, we get:
τ = 1.68 N * m
Therefore, the net torque acting on the rod is 1.68 N * m.