Question

aerslon 2 Part D (Rotational Dynamics and Oscillations) Problem D1: A disk-shaped pulley has mass M 4kg and radius R 0.5cm, It rotates freely on a horizontal axis. A block of mass m 2kg hangs by a string that is tightly wrapped around the pulley. Assume the system starts from rest. Moment of inertia of a disk is I Ma 4-0.5 5 2 M R 1) What is the acceleration of the block? 2) What is the angle velocity of the pulley 3s after block is released? 3) Using the conservation of mechanical energy, find the speed of the block after it has fallen 1.6m. DOX-Mag Mitimz 2435 m 4+Z 2) Page 4 of 4

Answer 1

Radius of pulley R= 0.5cm = 0.005m

Mass of pulley M_p= 4kg

Block of mass M_b = 2kg

Angular momentum of pulley is € = a/R

Moment of inertia of pulley is I = M_pR²/2

Let tension in string is T.

Formula for torque on pulley at corner point

₱ = TR

Formula for torque in terms of angular acceleration

₱ = I€ (here I is moment of inertia and

€ is angular acceleration)

As both torque are equal

So TR = I€

TR = M_pR^{​​​​​​2}×a/2R

T = M_pa/2

Now balance force in block

T is upward

M_bg is downward and T is upward

So M_bg -T = M_ba

T = M_bg -M_ba

As acceleration of block and pulley are equal

Equating T(tenaion) from both equation.

M_bg -M_ba = M_pa/2

2×10 - 2a = 4a/2

20 = 4a

a = 5m/s²

B) Angular momentum of pulley is

€ = a/r

€ = 5/0.005

€= 1000rad/s²

Angular velocity w at t=3s

Using formula

_W= €t

W= 1000×3

W = 3000rad/s

C)

Change in height is 1.6m

so change in potential energy is m_bgh

Change in kinetic energy

1/2m_bv² +1/2Iw²

Using energy conservation

1/2m_bv² +1/2Iw² =m_bgh

2/2v² + m_pr²w²/4 = 2×10×1.6 (I = mr²/2)

V² + r²(v/r)² =32 (w=v/r)

V² = 32/2

V= 4m/s