pendulum questions (1 Viewer)

[haboozin]

hey i have 2 mechanics questions.
whoever answers them, can you please not skip working?
explain why you did everything. Thanks!


1.
A light inextensible string of length 3l is threaded through a smooth vertical ring which is free to turn. the string carries a particle at each end. One particle A of mass m is at rest at a distance L below the ring. the other particle B of mass M is rotating in a horizontal circle whose centre is A. Find the angular velocity of B and find m in terms of M.



2. Diagram attached.
Two particles are connected by a light inextensible string which passes through a small hole with smooth edges in a smooth horsizontal table. One particle of mass m travels in a circle on the table with constant angular velocity w (small omega). The second particle of mass M travels in a circle with constant angular velocity of Q (big omega) on a smooth horizontal floor distance x below the table. the length of string on the table and below the table are l and L respectively and the length L makes an angle @ wiht the vertical

i. if the floor exerts a force N on the lower particle, show N = M(g - xQ²) state the maximum possible value of Q for the motion to continue as described. what happens if Q exceeds this value?
second part is easy ... when N = 0 and if it exerts it will leave the table

ii. by considering the tension force in the string, show L/l = m/M(w/Q)²

iv.
done this one already d/w about it, incase if others want some practice:

if the lower particle exerts zero force on the floor , show that the tension T in the string is given by T=MgL/x

 

[KFunk]

hey i have 2 mechanics questions.
whoever answers them, can you please not skip working?
explain why you did everything. Thanks!


1.
A light inextensible string of length 3l is threaded through a smooth vertical ring which is free to turn. the string carries a particle at each end. One particle A of mass m is at rest at a distance L below the ring. the other particle B of mass M is rotating in a horizontal circle whose centre is A. Find the angular velocity of B and find m in terms of M.

Let the ring be O. AOB is a Δ where ∠AOB = π/3 and ∠ABO = π/6 (since AO = l , BO = 2l and AB = √3l).

The ∑vertical forces on A = 0 , so T = mg (1) where T is the tension in the string AO = tension in BO.

The ∑ of vertical forces on B = 0, so Tsinπ/6 = Mg, T = 2Mg (2)

equating (1) with (2) yields m=2M


The ∑ of horizontal forces on B = Mω²r = Mω²√3l
so Tcosπ/6 = Mω²√3l , but T = 2Mg

(M√3)g = ω²(M√3)l

∴ ω = √(g/l)

Edit: Missed a few capital M's

 

[haboozin]

and the second question????????????

 

[noah]

for 2 i:

by drawring the forces accociated with particle B we can see that:

T(y) + N = Mg (1)

where T(y) is the vertical component of the Tension force T.

we can also see that

T(y) = T cos@ (2)

T(x) = T sin@ (3) and that:

T(x) = MrQ^2 (4) (centripital force)

Threrfore

T sin@ = MrQ^2 sub (3) into (4)

T = (MrQ^2)/sin@ (5)

T(y) = cot@ MrQ^2 (5) into (2)


from the diagram we can see that the triangle formed has sides x, L and r with an angle of @ between lengths x and L and a right angle between the lengths x and r. where r is the radius.

therefore

cot@ = x/r

and T(y) = xMQ^2 (6)

by subbing (6) into (1):

xMQ^2 + N = Mg

therefore:

N = M(g - xQ^2)


EDIT: sorry, particle A is that of mass m and particle B is that of mass M

 

[noah]

for 2 ii:

by drawring the forces in for particle A we can see that:

T = mlw^2 (7)

where T is the tension force which is the same the whole way through the string and therefore is the same for particle B. l is the radius.

from (4):

T(x) = MrQ^2

from the diagram we see that:

r = L sin@

and from (3):

T sin@ = T(x)

therefore:

T sin@ = ML sin@ Q^2

T = MLQ^2 (8)

By (7)/(8)

1 = (m/M)(l/L)(w/Q)^2

therefore:

L/l = (m/M)(w/Q)^2