Question

bead of mass rm slides smouthly frum point A to point B on a semicircular horizontal wire loop nf radius R Ir is attracted twward its starting point A by a farce F directly pruportional to its distancer lrom Λ lur Instance, y u can imagine an elastic string cunnecting m to A). When it reaches B the lorce toward Л is F. Nustencal values: F-2N,R,20cm, m 500 g. a) Calculate the wurk done against this furce when the particle is heing displaced from A to B b] Now the bead is given a very slight displaement (-initial velocity negligible) from B and released to Iet it go back to A. Calculale the speed and velocity it will have at mid point and at A. c] How lung would it uake to get back from B to A? Does this time depend on R? d) At what point the force dane by the wire is maximum

A bead of mass m slides smoothly from point A to point B on a semicircular horizontal wire loop of radius R. It is attracted toward its starting point A by a force F directly proportional to its distance r from A (for instance, you can imagine an elastic string connecting m to A). What it reaches B the force toward A is Fo. Numerical values: Fo = 2 N, R = 20 cm, m = 500 g.

1. Calculate the work done against this force when the particle is being displaced from A to B.
2. Now the bead is given a very slight displacement (initial velocity negligible) from B and released to let it go back to A. Calculate the speed and velocity it will have at mid point C, and at A.
3. How long would it take to get back from B to A? Does this time depend on R?
4. At what point is the force done by the wire at its maximum?

Answer 1

a)we know that the work done is given by force*displacement.

for a small work done dw=fds. ....1

it is given that force is directly proportional to distance r from A. so   or f=contant*r ot f=kr. substituting this in equation 1

x r

gives,

dw=krdr, integrating both sides give,

dw=| krdr

w=2kR².....2

we also know F₀=k2R ......3(as force is proportional to distance from A). substituting this in eqn 2 gives,

w=F₀R=0.4 J

b) we know that f=ma from Newton's law

so, kr=mdv/dt

multiplying both sides by v gives,

krv=mvdv/dt

gives, krdr/dt=mvdv/dt. (since v=dr/dt)

krdr=mvdv integrating both side gives

......................4

rdr = m | udu

substituting value of k from eqn 3 gives

$v=r\sqrt{Fo/2mR}$............5

to find velocity at point c we put r=...(since displacement from point C to A is this value from pythagoras theorem)

V2R

therefore, v_c=$\sqrt{FoR/m}$ =0.894m/s and

velocity at A v_a=$\sqrt{2}*\sqrt{FoR/m}$=1.26m/s

similarly to calculate speed we change the value of r in eqn 5 as,

speed at point c implies r=pi*R/2 as distance between C and B is pi*R/2, gives speed=0.992m/s

and speed at point A implies r=pi*R (circumference of semicircle) which gives speed=1.985m/s

c)time to reach from B to A,t=displacement/velocity.

t=2R/1.26=0.317seconds and it does depend on R

d)the force done by the wire is maximum at point B as the string is extended maximum at that point.