Two metal disks, one with radius R1 = 2.40 cm and mass M1 = 0.900 kg and the other with radius R2 = 4.95 cm and mass M2 = 1.50 kg , are welded together and mounted on a frictionless axis through their common center. A light string is wrapped around the edge of the smaller disk and a 1.55 kg block is suspended from the free end of the string.
What is the magnitude of the downward acceleration of the block after it is released?
Take the free fall acceleration to be 9.80m/s2.
Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk.
Take the free fall acceleration to be 9.80m/s2.
You can also use mechanical energy conservation
Delta E(mec) = E (f)-E (i) = 0 -> E (f)= E (i)
---> K (f) + U(f) = K (i) +U (i)
Since it is released from rest , intial linear speed Vi= 0, and angular speed w(i) =0, and when it touches the floor h= 0, and final potential energy U (f)= 0
So K (f) = U (i) ---> (I w^2)/2 + (mv^2) 2 = mgh
W=V/R ---> (I v^2)/R^2 + mv^2 = 2mgh ----> v = sqrt ( 2mgh/ (I/R^2+m))
From that equation, you can find its velocity before it touches the floor
To find tangential acceleration of tge circular motion of the disk =linear acceleration of the block , thus equation can give the answer
(Vf)^2-(Vi)^2 = 2 a ( y2-y1) where Vf is the velocity found above, Vi =0
y2-y1 = the given height or verical distance from the hanging block to the floor
a =(Vf)^2/ 2h
That way is for finding both velocity and acceleration of the object before it touches the floor
Or it will be far much simpler using net torque equation
RT = I *alpha
Multiplying both sides by R --->( R^2)T = I * R*alpha
a = R*alpha --->( R^2 )T = I*a ---> a =( R^2)*T / I
You have got tension T, and rotational inertia "I" in part a to plug in to get the acceleration
Your part b can be shortened by just replacing R1 with R2 in
a1 = mg/ (m + I/( R1)^2) -> a2= mg/(m+ I/(R2)^2)
Instead of repeating the same steps again
Well, I have to say, this situation is really "spinning" my head! But fear not, I am here to make it as amusing as possible.
(a) To find the magnitude of the downward acceleration of the block when the string is wrapped around the smaller disk, we can use the principle of conservation of angular momentum.
First, let's calculate the moment of inertia of the disk made by welding the two disks together. The moment of inertia of a disk is given by I = (1/2) * m * r^2, where m is the mass and r is the radius.
For the smaller disk:
I1 = (1/2) * M1 * R1^2
For the larger disk:
I2 = (1/2) * M2 * R2^2
Now, since the disks are welded together, their combined moment of inertia is simply the sum of their individual moments of inertia:
I_total = I1 + I2
Next, let's find the torque acting on the system. Torque is defined as T = I * α, where α is the angular acceleration. In this case, the torque is due to the weight of the block, so T = m * g * r, where m is the mass of the block, g is the acceleration due to gravity, and r is the radius of the smaller disk.
Finally, we can use Newton's second law for rotational motion, ΣT = I * α, to relate the torque and the angular acceleration:
m * g * r = I_total * α
Solving for α, we get:
α = (m * g * r) / I_total
Now, since the block is in free fall, the magnitude of the downward acceleration of the block is equal to the radius of the disk times the angular acceleration:
a = r * α
Substituting the values we know, we get:
a = r * ((m * g * r) / I_total)
Plug in the values for m, g, r, M1, R1, M2, and R2 and you'll find the answer.
(b) Now, let's move on to the situation where the string is wrapped around the larger disk. This time, the process is similar. We just need to calculate the moment of inertia and solve for the magnitude of the downward acceleration using the same equations.
I1 remains the same because it is solely determined by the smaller disk. However, the torque acting on the system changes to T = m * g * r2, where r2 is the radius of the larger disk.
Similarly, the acceleration can be found using the formula:
a = r2 * ((m * g * r2) / I_total)
Plug in the values for m, g, r2, M1, R1, M2, and R2, and voila! You've got your answer.
Hope these explanations have lightened up your physics problem a bit! Good luck with your calculations!
To find the magnitude of the downward acceleration of the block in both cases, we can use the principles of rotational motion and Newton's Second Law.
First, let's consider the case when the string is wrapped around the edge of the smaller disk.
1. Calculate the moment of inertia (I1) of the smaller disk:
The moment of inertia of a solid disk is given by the formula: I = (1/2) * M * R^2
Substituting the values: I1 = (1/2) * 0.900 kg * (0.0240 m)^2
2. Calculate the torque (τ1) exerted on the combined system:
The torque is given by the formula: τ = I * α
Since the system is initially at rest, the angular acceleration (α) is equal to the linear acceleration (a), and we can substitute the moment of inertia (I1) and the linear acceleration (a) into the formula to calculate the torque.
3. Calculate the tension force (T1) in the string:
The tension force in the string is related to the torque by the formula: τ = T * R, where R is the radius of the smaller disk. Rearranging the formula, we get T1 = τ1 / R1
4. Apply Newton's Second Law:
The net force acting on the system is the force due to the tension in the string (T1) minus the weight of the block (m * g), where g is the acceleration due to gravity. The net force is equal to the mass of the combined system (M1) multiplied by the acceleration (a). Using this information, we can write the equation: T1 - m * g = M1 * a
Now, we can solve this equation to find the magnitude of the downward acceleration (a).
Repeat the same steps for the case when the string is wrapped around the edge of the larger disk by substituting the values with the appropriate variables and calculating the torque (τ2), tension force (T2), and applying Newton's Second Law.
By following these steps, you should be able to calculate the magnitudes of the downward acceleration for both cases.
mass of first disk = M1 = 0.750 kg
radius of first disk = R1= 2.45 cm = 0.0245 m
mass of second disk = M2 = 1.55 kg
radius of second disk = R2= 5.10 cm = 0.0510 m
moment of inertia of first disk = I1= (0.5) M1 R12 = (0.5) (0.750) (0.0245)2 = 2.251 x 10-4 kgm2
moment of inertia of second disk = I2= (0.5) M2 R22 = (0.5) (1.55) (0.0510)2 = 2.016 x 10-3 kgm2
Total moment of inertia of the combination = I = I1 + I2 = 2.251 x 10-4+ 2.016 x 10-3= 2.241x 10-3 kgm2
mass of the block hanging = m = 1.60 kg
part a)
for the hanging block , the force equation is given as ::
mg - T = ma Eq-1
for the disk system , torque is given as
Torque = T R1
but torque = I \alpha
so I \alpha = T R1
T = Ia/R21 Eq-2
from equation 1 and 2
mg - Ia/R21 = ma
a = mg/ (m + I/R21)
inserting the value
a = (1.6 x 9.8) / (1.6 + (2.241 x 10-3)/(0.0245)2)
a = 2.94 m/s2
part b)
for the hanging block , the force equation is given as ::
mg - T = ma Eq-1
for the disk system , torque is given as
Torque = T R2
but torque = I \alpha
so I \alpha = T R2
T = Ia/R22 Eq-2
from equation 1 and 2
mg - Ia/R22 = ma
a = mg/ (m + I/R22)
inserting the value
a = (1.6 x 9.8) / (1.6 + (2.241 x 10-3)/(0.051)2)
a = 6.37 m/s2