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
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.
3 answers
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
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
a1 = mg/ (m + I/( R1)^2) -> a2= mg/(m+ I/(R2)^2)
Instead of repeating the same steps again