A horizontal rod of mass m = 20 kg and length L = 5 meters is fixed on the left end to a horizontal, frictionless axis. The other end is held by a thread attached to the ceiling. The rod's moment of inertia is mL2/3.(a) What is the force that the axis exerts on the rod? N(ai) What is the CW torque caused by the mass of the rod about the axis?
Question
A horizontal rod of mass m = 20 kg and length L = 5 meters is fixed on the left end to a horizontal, frictionless axis. The other end is held by a thread attached to the ceiling. The rod's moment of inertia is mL2/3.(a) What is the force that the axis exerts on the rod? N(ai) What is the CW torque caused by the mass of the rod about the axis?
Solution
(a) The force that the axis exerts on the rod can be found by considering the forces acting on the rod. The weight of the rod acts downwards at its center of mass, which is at a distance L/2 from the axis. This force is given by F = mg, where g is the acceleration due to gravity. The tension in the thread and the force exerted by the axis must balance this weight. Since the thread is vertical, its tension does not contribute to the vertical forces. Therefore, the force exerted by the axis is equal to the weight of the rod, F = mg = 20 kg * 9.8 m/s^2 = 196 N.
(ai) The clockwise torque caused by the mass of the rod about the axis can be found by considering the moment of the weight about the axis. The weight acts at the center of mass of the rod, which is a distance L/2 from the axis. Therefore, the torque is given by τ = rFsinθ, where r is the distance from the axis, F is the force, and θ is the angle between r and F. Since r and F are perpendicular, sinθ = 1. Therefore, τ = (L/2)mg = (5 m/2) * 20 kg * 9.8 m/s^2 = 490 N*m.
Similar Questions
A uniform rod of mass 3 kg and length 1 m is free to rotate about an axis which is perpendicular to the rod and 0.3 m from its centre of gravity. Use the parallel axis theorem to find the moment of inertia of the rod about this axis
A non-uniform rod of length L and mass M is pivoted about one end. The moment of inertia of this rod about its pivoted end is . Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass 2M. What is the moment of inertia of this configuration about the pivot?
A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?Select one:a.5.88 rad/s2b.3.70 rad/s2c.4.90 rad/s2d.7.35 rad/s2
A 3.0-m rod is pivoted about its left end. A force of 20 N is applied perpendicular to the rod at a distance of1.2 m from the pivot causing a ccw torque, and a force of 4.8 N is applied at the end of the rod 3.0 m from the pivot. The 4.8-N force is at an angle of 30o to the rod and causes a cw torque. What is the net torque about the pivot? (Take ccw as positive.)Select one:a.-16.8 N·mb.31.2 N·mc.-31.2 N·md.16.8 N·me.9.6 N·m
The three objects shown in the figure above have masses given byA: 220. gB: 260. gC: 60.0 g.They are connected by massless, rigid rods. Find the moment of inertia about an axis that passes through mass B and is perpendicular to the page. 0.0593 kg m2 0.0280 kg m2 0.00540 kg m2 0.00280 kg m2
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.