A 9.35 kg block is being lowered down a ramp, inclined at θ = 39 degrees, by the 4.05 kg pulley. The pulley has a radius of 45.0 cm ?

Question

and you can assume there is no friction between the block and the ramp. What is the acceleration of the block?

Answer 1

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Answer: a = 5.07 m/s²

What accelerates the block is its weight, more precisely the component of weight in direction of slope

F = mg sin θ

What opposes this force is the inertia of the block plus inertia of the pulley. Problem is that the pulley is rotating so we need to reduce its angular inertia to an equivalent tangential linear force.

2nd Law for rotation

T = I 𝛼

T is torque

I is moment of inertia

𝛼 is angular acceleration

The torque T that accelerates the pulley is tangential force Ft acting on pulley radius R

T = Ft R

Moment of inertia for disc of mass M

I = M R² / 2

𝛼 = a / R

"a" is the linear acceleration we are searching for.

2nd Law again

T = I 𝛼

Ft R = (M R² / 2) (a / R)

Ft = M a  / 2 ......... (*)

2nd Law for the system

F = ma + Ft.     ........... (**)

mg sin θ = ma + Ma/2

mg sin θ = a(m + M/2)

a = mg sin θ / (m + M/2) ........ (***)

a = 9.35*9.80*sin 39° / (9.35 + 4.05/2)

a = 5.07 m/s²

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Slightly different approach. Reduced mass.

Observe the equation (*)

We see that when we accelerate a disc of mass M it is equivalent to linear acceleration of point mass (M/2) sitting on disc's perimeter.

We could write 2nd Law for the system (**) as

a = ΣF / Σm

ΣF = sum of active forces

Σm = sum of masses

a = F / (m + M/2) = mg sin θ / (m + M/2)

where we replaced the influence of disc with reduced mass M/2

This is the result we had in (***)

Answer 2

acceleration a = mb*g*sin 39/(mb+mp/2) 

a = 9.35*9.806*0.629/(9.35+4.05/2) = 5.070 m/sec^2