Centre of Mass 4 Question 10

11. A uniform thin rod of mass M and length L is standing vertically along the Y-axis on a smooth horizontal surface, with its lower end at the origin (0,0). A slight disturbance at t=0 causes the lower end to slip on the smooth surface along the positive X-axis, and the rod starts falling. (1993,1+5M) (a) What is the path followed by the centre of mass of the rod during its fall?

(b) Find the equation of the trajectory of a point on the rod located at a distance r from the lower end. What is the shape of the path of this point?

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Answer:

Correct Answer: 11. (a) Straight line

(b) x2L2r+y2r2=1, ellipse

Solution:

  1. (a) Since, only two forces are acting on the rod, its weight Mg (vertically downwards) and a normal reaction N at point of contact B (vertically upwards).

(a)

(b) No horizontal force is acting on the rod (surface is smooth).

Therefore, CM will fall vertically downwards towards negative Y-axis i.e. the path of CM is a straight line.

(b) Refer figure (b). We have to find the trajectory of a point P(x,y) at a distance r from end B.

CB=L/2OB=(L/2)cosθ;MB=rcosθx=OBMB=cosθ(L/2r) or cosθ=x(L/2)r

Similarly, y=rsinθ

or

sinθ=yr

Squaring and adding Eqs. (i) and (ii), we get

sin2θ+cos2θ=x2(L/2)r2+y2r2x2(L/2)r2+y2r2=1

This is an equation of an ellipse. Hence, path of point P is an ellipse whose equation is given by (iii).

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