A uniform cylindrical spool of mass M and radius R unwinds an essentially
massless rope under the weight of a mass m. If R = 12 cm, M = 400 gm and m = 50 gm, find the speed of m after it has descended 50 cm starting from rest.
Solve the problem twice: once using Newton's laws for torques, and once by application of energy conservation principles.
Ok, while the first part may have confused me, I may have gotten close to the answer:
I(spool) = 0.5MR^2
= 0.00288 kg m^2
Torque = R *(Fg + Ft - Ft)sin(angle)
= R * (mg)sin(90)
= Rmg
= -5.88 x 10^-2 N-m (I converted everything to meters and kg)
So then:
Torque = I * alpha
alpha = Torque/I
But since we want to find the speed of mass m after it has fallen 50 cm, alpha is kind of a useless value to us. So I used the conversion that
a(tangential) = alpha * radius
alpha = a(t)/radius
Torque/I = a(t)/radius
a(t) = radius * Torque/I
= -20.4 m/s
I then solved for the v(final) of mass m after 50 cm of falling using the equation:
v(f) = sqrt(vi^2 + 2*a(t) * delta(x))
and got -4.52 m/s
However, I am really confused about how to even approach the problem using energy conservation principles. How would you go about doing that at all?
Well, my circus of thoughts may have some amusement for you. Let's unravel this problem using energy conservation principles!
When mass m descends, its potential energy is converted into kinetic energy. We can start by calculating the initial potential energy.
The potential energy at the start is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height from which m descends.
In this case, the height h is 50 cm, which is 0.5 meters. Plugging in the values, we get potential energy at the start as mgh = (0.05 kg) * (9.8 m/s^2) * (0.5 m) = 0.245 J.
Now, let's find the final kinetic energy of mass m. Since it started from rest, the initial kinetic energy is zero. The final kinetic energy is given by (1/2)mv^2, where v is the final velocity we need to find.
Since energy is conserved, the initial potential energy must be equal to the final kinetic energy: 0.245 J = (1/2)(0.05 kg)v^2.
Let's solve this equation to find v:
0.245 J = (1/2)(0.05 kg)v^2
0.49 = (0.05 kg)v^2
v^2 = 9.8 m^2/s^2
v ≈ 3.13 m/s
So, the speed of mass m after it has descended 50 cm is approximately 3.13 m/s when using energy conservation principles.
Hope that brightened up the problem-solving journey a bit!
To solve the problem using energy conservation principles, we need to consider the change in potential energy and the change in kinetic energy of the system.
1. First, let's calculate the change in potential energy. The mass m has descended 50 cm, so the change in height is Δh = 0.5 m. The potential energy change is given by ΔPE = mgh, where g is the acceleration due to gravity (9.8 m/s²).
ΔPE = (0.05 kg)(9.8 m/s²)(0.5 m)
= 0.245 J (Joules)
2. Next, let's calculate the change in kinetic energy. The initial kinetic energy of mass m is zero because it starts from rest. The final kinetic energy is given by KE = 0.5mv², where v is the final velocity.
ΔKE = 0.5mv² - 0.5m(0)²
= 0.5mv²
3. According to the law of conservation of energy, the change in potential energy must be equal to the change in kinetic energy.
ΔPE = ΔKE
0.245 J = 0.5mv²
Substituting the given values:
0.245 J = 0.5(0.05 kg)v²
v² = (0.245 J) / (0.5)(0.05 kg)
v² = 9.8 m²/s²
v = √(9.8 m²/s²)
v ≈ 3.13 m/s
Therefore, the speed of mass m after it has descended 50 cm starting from rest is approximately 3.13 m/s when using the principles of energy conservation.
To solve the problem using energy conservation principles, we need to consider the change in potential energy and the change in kinetic energy as the mass descends.
1. Potential Energy Approach:
The potential energy of the mass m at a height h is given by the equation: PE = m * g * h, where g is the acceleration due to gravity.
Since the mass m descends 50 cm, we can calculate the initial potential energy (PI) and the final potential energy (PF).
PI = m * g * initial height
= 0.05 * 9.8 * 0.5
= 0.245 J
PF = m * g * final height
= 0.05 * 9.8 * 0.0
= 0 J (since the mass has descended all the way)
The change in potential energy is given by ΔPE = PF - PI
= 0 - 0.245
= -0.245 J
According to the principle of conservation of energy, this change in potential energy is equal to the change in kinetic energy.
2. Kinetic Energy Approach:
The kinetic energy of the mass m is given by the equation: KE = 0.5 * m * v^2, where v is the velocity.
Since the mass starts from rest, the initial kinetic energy (KI) is 0 J.
The final kinetic energy (KF) can be calculated by comparing it to the change in potential energy:
ΔPE = KF - KI
-0.245 = KF - 0
Therefore, KF = -0.245 J.
Now, we can substitute this value of KF into the kinetic energy equation and solve for v:
-0.245 = 0.5 * 0.05 * v^2
v^2 = -0.245 / (0.5 * 0.05)
v^2 = -9.8
v = √(-9.8)
v = √(-1) * √(9.8)
v = i√9.8 (taking the imaginary square root)
Since the velocity cannot be imaginary in this context, it indicates that there is no solution for the speed of m after it has descended 50 cm starting from rest using the energy conservation approach.
Hence, the Newton's laws for torques approach gives the correct answer of -4.52 m/s.
Using energy principles,
(KE=kinetic energy, PE=gravitational potential energy)
KE+PE=constant, so
KEi+PEi = KEf+PEf
KEi=0 (both are stationary)
KEf=(1/2)Iω^2+(1/2)mv^2
PEi=0
PEf=0+mgh (weight descended)
Equate energies:
0 = (1/2)Iω^2+(1/2)mv^2 + 0+mgh
(note that h=-0.50
Everything is given, except v and ω
However, v and ω are related by
v=Rω, or ω=v/R
so we can solve for v in:
(1/2)[(1/2)MR^2](v/R)^2+(1/2)mv^2 + mgh = 0
Using
m=.05 kg,
M=0.4 kg
R=0.12 m
h=-0.5 m
g=9.81 m/s²
you will get v
(between 1.4 and 1.5 m/s)