Question

Consider two uniform circular disks one with mass m1 = 1kg, and radius R1 = 0.1m; and the other with mass m2 = 2kg, and radius R2 = 0.2m mounted together so that they can be rotated around a thin mass-less and friction-less horizontal rod through their center. A light string is wrapping around the edge of the smaller disk, and a block with mass m3 = 3 kg is suspended from the free end of the string. If the block is released from rest at a distance of 2m above the floor, show that (1) the translational acceleration of the block: a = 5.35 m/s^2 ; (2) The tension on the sting: T = 13.35 N; determine (3) the speed of the block just before it hits the floor; (4) Work done [in J] (i) by Gravity on the block; (ii) by the Tension of the String on the Disks. [Moment of Inertia of a Uniform Circular Disk around an axis through its center: I = (1/2)MR^2

Answer 1

Answer:

(1) a = 5.35 m/s²

(2) T = 13.35 N

(3) v = 4.62 m/s

(4)(i) W = 58.8 J

(4)(ii) W = 26.7 J

Explanation:

(1) Sum of forces on the block in the downward direction:

∑F = ma

m₃g − T = m₃a

Sum of torques on the disks:

∑τ = Iα

TR₁ = (I₁ + I₂) (a / R₁)

T = (½ m₁R₁² + ½ m₂R₂²) a / R₁²

T = ½ (m₁ + m₂ (R₂/R₁)²) a

Substitute:

m₃g − ½ (m₁ + m₂ (R₂/R₁)²) a = m₃a

m₃g = ½ (m₁ + m₂ (R₂/R₁)² + 2m₃) a

a = 2m₃g / (m₁ + m₂ (R₂/R₁)² + 2m₃)

Plug in values:

a = 2 (3) (9.8) / (1 + 2 (0.2/0.1)² + 2(3))

a = 3.92 m/s²

This does not match the expected answer. However, if m₂ = 1 kg instead of 2 kg, then:

a = 2 (3) (9.8) / (1 + 1 (0.2/0.1)² + 2(3))

a = 5.35 m/s²

(2) Use answer from (1) to find the tension.

m₃g − T = m₃a

T = m₃ (g − a)

T = (3) (9.8 − 5.35)

T = 13.35 N

(3) Given:

s = 2 m

u = 0 m/s

a = 5.35 m/s²

Find: v

v² = u² + 2as

v² = (0)² + 2 (5.35) (2)

v = 4.62 m/s

(4)(i) Work = force × distance

W = mgh

W = (3) (9.8) (2)

W = 58.8 J

(4)(ii) Work = torque × angular displacement

W = τθ = Th

W = (13.35) (2)

W = 26.7 J