Analyzing a Pivoted Light Rod with Concentrated Masses

[The Futur]

Homework Statement

(b) Fig. Q.5 shows a light rod pivoted at its centre. There are concentrated masses of 0.6 kg located at each end, 0.3 m from the centre of rotation The rod is rotated by a falling mass suspended from a cord wrapped around the centre spindle of diameter 50 mm. If the rod and masses have an angular acceleration of 5 rad/s2,

Determine:

(i) The moment of inertia of the assembly, neglecting that of rod and spindle

(ii) The torque on the spindle (assuming friction torque is negligible)

(iii) The tension in the cord

(iv) The angular velocity 5 seconds after release of the mass

I attached the fig5

NB: i don't want the answers i just need, some one to explain me how to do it. and witch formula i need to use.

 

[nvn]

The Futur: You deleted two important parts of the required template. The PF rules state, you must list relevant equations yourself, and you must show your work. Then someone might check your math.

 

[Mene]

I would approach this problem by first identifying the relevant physical principles and equations to use. In this case, the key concepts are rotational motion, torque, and moment of inertia.

(i) To determine the moment of inertia of the assembly, we can use the equation I = Σmr^2, where I is the moment of inertia, Σm is the sum of the masses, and r is the distance from the axis of rotation. In this case, we need to calculate the moment of inertia of the two masses located at each end of the rod. Since the masses are concentrated, we can treat them as point masses and use the formula I = mr^2. Plugging in the values, we get I = (0.6 kg)(0.3 m)^2 + (0.6 kg)(0.3 m)^2 = 0.108 kg m^2. This is the moment of inertia of the assembly, neglecting the rod and spindle.

(ii) The torque on the spindle can be calculated using the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, we know the moment of inertia from part (i) and the angular acceleration given in the problem. Plugging in the values, we get τ = (0.108 kg m^2)(5 rad/s^2) = 0.54 Nm. This is the torque on the spindle, assuming friction torque is negligible.

(iii) To determine the tension in the cord, we can use the equation τ = Fr, where τ is the torque, F is the tension force, and r is the radius of the spindle. We know the torque from part (ii) and the radius of the spindle is given as 50 mm, or 0.05 m. Plugging in the values, we get F = 0.54 Nm / 0.05 m = 10.8 N. This is the tension in the cord.

(iv) Finally, to calculate the angular velocity 5 seconds after the mass is released, we can use the equation ω = ω0 + αt, where ω is the final angular velocity, ω0 is the initial angular velocity (which is 0 in this case), α is the angular acceleration, and t is the time. Plugging in the values, we get ω