A rod of weight w is supported by two parallel knife

A mass m moves in a circle on a smooth horizontal plane with velocity v 0 at a radius R 0. The mass is attached to a string which passes through a smooth hole in the plane as shown. the tension in the string is increased gradually and finally m moves in a circle of radius $$$$. The final value of the kinetic energy is

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Learning Objectives By the end of this section, you will be able to: • Identify and analyze static equilibrium situations • Set up a free-body diagram for an extended object in static equilibrium • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Problem-Solving Strategy: Static Equilibrium • Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly. • Set up a free-body diagram for the object. (a) Choose the xy-reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x– and y-directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction i...

As the weight w balances the normal reaction. So, w = N 1 ​ + N 2 ​ ...(i) Now balancing torque about the COM, i.e., anti-clockwise momentum = clockwise momentum ⇒ N 1 ​ x = N 2 ​ ( d − x ) Putting the value of N 2 ​ from Eq. (i), we get N 1 ​ x = ( w − N 1 ​ ) ( d − x ) ⇒ N 1 ​ x = w d − w x − N 1 ​ d + N 1 ​ x ⇒ N 1 ​ D = w ( d − x ) ⇒ N 1 ​ = d w ( d − x ) ​ Views: 5,875 x simultaneously with speeds 16 m / sec and 12 m / sec respectively. In order to win the race they increase their speeds at the rate 1 m / s 2 and 2 m / s 2 respectively. Unfortunately both reach the final point y at the same instant of time. 8. If the time required to cover the distance x y is t then : A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is.. And on B is...... Updated On Mar 14, 2023 Topic System of Particles and Rotational Motion Subject Physics Class Class 11 Answer Type Text solution:1 Video solution: 9 Upvotes 1054 Avg. Video Duration 4 min

Physics A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d-from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d-from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

Solution: Given situation is shown in figure. N 1 ​ = Normal reaction on A N 2 ​ = Normal reaction on B W = Weight of the rod In vertical equilibrium, N 1 ​ + N 2 ​ = W ...... ( i ) Torque balance about centre of mass of the rod, N 1 ​ x = N 2 ​ ( d − x ) Putting value of N 2 ​ from equation (i) N 1 ​ x = ( W − N 1 ​ ) ( d − x ) ⇒ N 1 ​ x = W d − W x − N 1 ​ d + N 1 ​ x ⇒ N 1 ​ d = W ( d − x ) ∴ N 1 ​ = d W ( d − x ) ​

A rod of weight `w` is supported by two parallel knife edges `A` and `B` and is in equilibrium in a horizontal position. The knives are at a distance `d` from each other. The centre of mass of the rod is at a distance `x` from `A`. A. the normal reaction at `A` is `(wx)/(d)` B. the normal reaction at `A` is `(w(d - x))/(d)` C. the normal reaction ar `B` is `(wx)/(d)` D. the normal reaction at `B` is `(w(d - x))/(d)`.