Answer:
Explanation:
Given that,
His friend design has a turnable disk of radius 1.5m
R = 1.5m
The mass is twice the fully loaded elevator.
Let the mass of the full loaded elevator be M
Then, mass of the turn able
Mt = ½M
Radius of the disk that serves as a vertical pulley is ¼ radius of turntable and 1/16 of the mass.
Mass of pulley is
Mp = 1 / 16 × Mt
Mp = 1 / 16 × M / 2
Mp = M / 32
Also, radius of pulley
Rp = ¼ × R = ¼ × 1.5
Rp = 0.375m
The moment of inertia of the disk of a ring is
I = ½MR²
To calculate the moment of the turntable, we can use the formula
I_t = ½Mt•R²
I_t = ½ × ½M × 1.5²
I_t = 0.5625 M
I_t = 9M / 16
Also, the moment of inertia of the vertical pulley
I_p = ½Mp•Rp
I_p = ½ × (M/16) × 0.375
I_p = 0.01171875 M
I_p = 3M / 256
Let assume that, the tension in the cable between the pulley and the elevator is T1 and Let T2 be the tension between the turntable and the pulley
So, applying newton second law of motion,
For the elevator
Fnet = ma
Mg - T1 = Ma
a = (Mg-T1) / M
For vertical pulley,
The torque is given as
τ_p = I_p × α_p = (T2—T1)•r
τ_p = 3M/256 × α_p = (T2-T1)•r
For turntable
The torque is given as
τ_t = I_t × α_t = T2•r
τ_t = 9M/16 × α_t = T2•r
So, the torque are equal
τ_t = τ_p
9M/16 × α_t = 3M/256 × α_p
M cancel out
9•α_t / 16 = 3•α_p / 256
Cross multiply
9•α_t × 256 = 3•α_p × 16
Divide both sides by 48
48•α_t = α_p
α_t = α_p / 48
Then, from,
τ_t = 9M/16 × α_t = T2•r
T2•r = 9M / 16 × α_p / 48
T2•r = 3Mα_p / 16
Also, from
τ_p = 3M/256 × α_p = (T2-T1)•r
3M/256 × α_p = T2•r - T1•r
T1•r = T2•r - 3M/256 × α_p
T1•r = 3Mα_p / 16 - 3M/256 × α_p
T1•r = 3Mα_p / 16 - 3Mα_p/256
T1•r = 45Mα_p / 256
T1 = 45Mα_p / 256R
Then, from
a = (Mg-T1) / M
a = Mg - (45Mα_p / 256R) / M
a = g - 45α_p / 256
From the final answer, it is show that the acceleration is always less than acceleration due to gravity due to the subtraction of 45α_p / 256 from g
