1. Physicists like to make things simple to begin with their approximations. Assuming that we are referring to Uniform Circular Motion (UCM) for short hereafter, it means a particle must be travelling in a circular path at constant SPEED. The velocity however of the particle must be changing.
This follows from Newton's first law of motion which requires that a force must be acting on the particle if its velocity is changing. Even if the magnitude or size of the force is constant but only the direction is changing, the VELOCITY is changing. Therefore a force MUST be acting on the particle.
This force is always acting on the body and is directed towards the centre of the circular path. It is therefore called the CENTRIPETAL FORCE (towards the center).
Circular motion is NOT equilibrium motion. There must be a resultant force and this resultant force that must be acting towards the centre is usually provided by forces such as tension in a cable, gravitational force, electrostatic force of attraction and so on.
In numerical problems, one looks for what provides this centripetal force and sets up a few equations that could be solved for finding out the unknowns.
2. This type of circular motion is typically classfied as a conical pendulum or motion on a horizontal circle problem.
a) Let the angle the rider makes with the vertical be = Ø. The forces acting on the RIDER would be the tension in the rope along the rope upwards, the weight vertically downward and the centrifugal force m(A + lsinØ)w2
Normally one does not speak about the centrifugal force because it is not a real force we can use to describe the motion but when you speak of the RIDER, s/he experiences an outward force just like you and I would when we negotiate a turn.
b) Resolving the tension into vertical and horizontal components, you'll have
TcosØ = mg
and
TsinØ = m(A + lsinØ)w2
Dividing the two equations
(A + lsinØ)w2/g = tanØ
c) The linear velocity
v = r.w
Using the work energy theorem,
work done W = change in rotational kinetic energy
W = (1/2)Iw2
where I is the rotational inertia given by
I = mr2
Here it will be SIX times the above value where
r = A + lsinØ
and w = v/r
3. This is a typical problem on S.H.M. We'd be using some basic equations here.
You're given the time period T. Therefore you can find the angular velocity w.
By definition
T = 2(pi)/w and making w the subject of the equation, you'll have
w = 2(pi)/T
The maximum acceleration is given by
a = w2A
where you've converted 10 mm to 0.01 m
To find the acceleration at a position x = 5 mm from the equilibrium position you use the formula
a = -w2x
(The negative sign has been included simply to represent the fact that the accelerartion is directed towards the mean position. Don't forget once again 5 mm has been converted to 0.005 m)
The force at this point can be calculated using Newton's second law
F = ma
4. The fact that you don't have friction on the surface is a boon in this problem because it gives you less forces to contend with.
Do you know why objects become easier to push along an inclined plane? The reason is only a part of the weight acts along the inclined plane. In this case it would be only part of 14 N, actually 14sin(40) = 9 N, which acts ALONG the plane. The perpendicular component does not affect the spring or the block because there is no friction.
Also if an object weight 9 N then its mass 'm' must be 0.92 kg (taking g = 9.8 ms-2, since weight W = mg).
To solve the first part of the problem about how far from the top of the incline will the block stop, you must find the extension. Using Hooke's law
F = kx you'll have
Since the spring already has an unstretched length of 0.45 m, the total distance from the top will be = 0.45 + 0.075 = 0.525 m
For the second part of the problem you would use the standard formula
T = 2(pi)[m/k]1/2
5. The centripetal force can be calculated using the formula
a) F = m(v)2/r
= (55)(2.5)2/(1.2)
b) The guy must exert this force of 287 N, which is what makes her to rotate in this circular path.
c) Her rotational inertial is
I = mr2
d) Her angular momentum is
L = Iw
To calculate her angular velocity w you can use the formula
v = rw
w = v/r
Subsituting this in L above you'll have
L = (19.8)(2.08)
From the law of conservation of angular momentum, in the absence of an external torque the total angular momentum must remain constant.
First let's find the new moment of inertia which I shall call 'i'.
Using the definition
i = mR2
If W is the new angular velocity, then
L = iW
giving you
W = 8.32 rads-1
which is much faster rate of rotation as you might have accepted.
6. a) The rotational inertia of the wheel is
I = mr2
b) The linear acceleration of the falling mass can be determined using the formula
mgh = 1/2(m)(v)2
giving you a velocity of
v = (2gh)0.5
Also
v = u + at
giving you
a = 4.7 ms-2
c) The angular acceleration can be found using the equation
W = w + (alpha)t
where W is the final angular velocity and w is the initial angular velocity (which is zero)
To find W we can use the formula
v = rW
therefore
W = 28 rads-1
Therefore
28 = 0 + (alpha)(3)
d) The torque
(Tau) = (I)(alpha)
e) The tension (F) in the rope can be calculated using the formula
(Tau) = F (perpendicular distance moved)
giving you the tension as F = 11.6 N
7. a) The average linear velocity of the ball
average velocity = distance/time
b) The final velocity can be calculated using the formula
s = 1/2(u+v)t
c) The accleration of the ball is given by
a = v - u/t
d) The potential energy at the top of the incline is
P.E. = mgh
= (10)(9.8)(4)
e) The total energy at the bottom must also be = 392 J
Of this the translational k.e. is
= 1/2(m)(v)2
Therefore the ramaining energy
8. a) The distance travelled is
s = 1/2(u+v)t
b) The angular distance
theta = arc length/radius
= cicumference/radius
c) The angular velocity
omega = v/r
Note you've converted the diameter to radius by dividing by two and the converted this to meters.
d) The linear acceleration is
a = v - u/t
e) The angular acceleration is
alpha = w/t
w2 = gtanØ/(A + lsinØ)
w = [gtanØ/(A + lsinØ)]1/2
= (A + lsinØ).w
= 2(3.14)/0.4
= 15.7 rads-1
= (15.7)2(0.01 m)
= 2.5 ms-2 (2 sig. figs.)
= -(15.7)2(0.005)
= 1.2 ms-2 (2 s.f.)
= (0.005)(1.2)
= 0.006 N or 6 x 10-3 N
9 = 120(x)
giving you a solution for x = 0.075 m
= 2(3.14)[0.92/120]1/2
= 0.55 s
= 287 N (3 sig. figs.)
= (55)(0.6)2
= 19.8 kgm2 (3 s.f.)
= (2.5)/(1.2)
= 2.08 rads-1 (3 s.f.)
= 41.2 kgm2s-2
= (55)(0.3)2
= 4.95 kgm2
41.2 = 4.95(W)
= (50)(0.5)2
= 12.5 kgm2
= (2x9.8x10)0.5
= 14 ms-1
14 = 0 + (a)(3)
[since the mass falls from rest the initial velocity u = 0]
14 = (0.5)W
(alpha) = 9.3 rads-2
= (12.5)(9.3)
= 116 Nm
116 = F (10)
= 12/3
= 4 ms-1
12 = 1/2(0 + v)(3)
giving you
v = 8 ms-1
= (8 - 0)/(3)
= 2.7 ms-2
= 392 J
= 1/2(10)(8)2
= 320 J
= 392 - 320
= 72 J
must be the rotational kinetic energy.
= 1/2(0 + 20)9
= 90 m
= 90/0.4
= 225 radians
= 50 rad/s
= v/t since u = 0
= 20/9
= 2.2 ms-2
= 50/9
= 5.6 rads-2