Q#25
A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called radius of curvature of the curve at the point.
Answer
You may remember such problems in projectile motion where the horizontal speed remains constant throughout its motion = u.cosθ. In this problem also near the highest position, same will be the speed, i.e. v = u.cosθ
If we consider a small part of its path as to be a circular arc, then the particle is moving with a uniform circular motion having this speed.
Take its mass as 'm', then inward force on this particle in circular motion is its weight, 'mg'. Now it can be related with its speed as,
mg = m(u.cosθ)²/R (Where R is the radius of curvature)
R = (u.cosθ)²/g
R = u²cos²θ/g
Q#26
What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal?
Answer:
Let us draw a diagram as below,
Take the speed of the projectile at the given point be 'v'.
Since the horizontal velocity remains constant, so
vcosθ/2 = ucosθ
v = u cosθ/(cosθ/2)
If we take its motion at the instant as uniform circular motion, we also need to know the centripetal force at the instant. It is provided by the component of the weight of the particle along the perpendicular to the speed v.
P = mgcosθ/2
Now we can relate v and P as,
P = mv²/R
mgcosθ/2 = mu²cos²θ/R(cos²θ/2)
R = u²cos²θ/g cos³(θ/2)
Q#27
A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is µ. The block is given an initial speed v0. As a function of the speed v write (a) the normal force by the wall on the block, (b)) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration {dv/dt = vdv/ds} to obtain the speed of the block after one revolution.
Answer:
Since the circular motion in this problem is not a uniform one, the block will have both the tangential and radial accelerations.
(a) Radial acceleration = v²/R (inward)
It is provided by the Normal force 'N' by the wall on the block.
So normal force by the wall on the block N = mv²/R
(b) The frictional force by the wall on the block,
F = µN
F = µmv²/R
(c) The tangential acceleration of the block is given by a = F/m.
Force of retardation on the block is frictional force 'F'
Since it is retardation, so we write
a = -F/m = -µmv²/mR = -µv²/R
(d) Since a = vdv/ds
Speed after one revolution (Covering a distance s = 2πR) is given by,
∫dv = ∫(a/v)ds = ∫(-µv²/vR)ds = -µv/R∫ds
∫(1/v)dv = (-µ/R)∫ds
{We integtrate dv from v0 to v and ds from 0 to 2πR for one revolution}
[ln v – lnv0] = (-µ/R)[2πR – 0] = -2πµ
ln (v/v0) = -2πµ
v/v0 = e–2πµ
v = v0e-2πµ
Q#28
A table with a smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R (figure 7-E3). A smooth groove AB of length L (<<R) is made on the surface of the table. The groove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.
Answer:
Since L<<R, we assume that force on the particle in the groove is constant from A to B.
Radial acceleration of the particle = ω²R
Component of acceleration along groove = ω²R cosθ
Let it take time t to cover AB, using s = ut + ½at²
L = 0 + ½ ω²R cosθt² (Initial velocity u = 0)
t² = 2L/ω²Rcosθ
t = √{2L/ω²Rcosθ}
Q#29
A car moving at a speed of 36 km /hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is µ = 0.58. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.
Answer:
Mass of the block, m = 100 g = 0.10 kg
v= 36 kmph =36000/3600 m/s =10 m/s
R = 50 m
(a) Normal force on the block by the plate = mv²/R
= 0.10 x 10²/50
= 0.20 N
(b) As the plate is turned, say by an angle θ, the normal force on the block by the plate decreases to a component of original normal force = mv²/Rcosθ N
Force of friction that will resist its sliding on the block will be
= µmv²/Rcosθ
While the force trying to slide it is = mv²/Rsinθ
At the point when the block will just start sliding, these two forces will be equal.
mv²/Rsinθ = µmv²/Rcosθ
tanθ = µ =0.58
θ = 30°
Q#30
A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure 7-E5). A smooth pully of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string over the pulley. Initially, the masses are held by a person with the strings along the outward radius and then the system is released from the rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.
Answer:
The system experiences an outward acceleration
= ω²R
Let tension in the string be T, m1 = m, m2 = 2m
If 'a' be the initial acceleration, For the first block,
T – mω²R = ma,
And for the second block,
2mω²R – T = 2ma,
Eliminating T in these two equations, we get
2mω²R – ma – mω²R = 2ma
3ma = mω²R
3a = ω²R
a = ω²R/3
Put the value of 'a' in first equation,
T = mω²R + mω²R/3
T = 4mω²R/3
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