A thin rod is pivoted (has an axis) at its end. Its mass 2 kg and its length is

Question

A thin rod is pivoted (has an axis) at its end. Its mass 2 kg and its length is 5 m. It rests on a smooth, frictionless surface. a) What is its moment of Inertia? (formula I = (1/3) ML) b) A force of 200 N act at an angle of 60 degree to the direction of the rod at its end. What is the torque from this force on end of the Rod? Include the direction of the torque (into or out of page?) c) What would be the angular acceleration of the rod at this instant? A cord is wrapped around the rim of a solid uniform wheel 0.200 m in radius and of mass 8.60 kg. A steady horizontal pull of 50.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. Compute the linear acceleration of the part of the cord that has already been pulled off the wheel. The moment of Inertia of the wheel is 1/2 MR^2. The value of the constant "g" is given by 9.8 m sec^2. A lever (Class 2) has a load of 306 gm at 30 cm from its axis. Its lever arm is uniform and one meter long with a mass of 102 grams. A lifting force will act on its far end (1 meter from the axis) a) On the diagram, Show ALL forces on lever arm (the lever arm is the Free Body) b) Write Equilibrium equations for the sum of Forces and the sum torques for the system. Assume an axis at the pivot, far left on the diagram. Sum of Forces eq:___ Sum of Torques____ c) Solve for the lifting force, F (units?)

Explanation / Answer

(a)

Moment of inertia:

I = (1/3)ML2 = (1/3)*(2 kg)*(5 m)2 = 16.67 kg.m2 = 16.7 kg.m2

(b)

magnitude of the torque:

t = rFsin60o = 5*200*sin60o = 866 N.m, into the page.

(c)

angular acceleration:

alpha = t/I = 866 / 16.67 = 51.95 rad/s2

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