Question

In the figure, two 5.60 kg blocks are connected by a massless string over a pulley of radius 2.20 cm and rotational inertia 7.40 times 10^-4 kg-m^2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 1.00 rad in 179 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley's angular acceleration, (b) the magnitude of either block's acceleration, (c) string tension T_1, and (d) string tension T_2? Assume free-fall acceleration to be equal to 9.81 m/s^2. Number Units Number Units Number Units Number Units

Answer 1

Moment of inertia of pulley is given by, I = (1/2) M R²

(7.4 x 10^-4 kg.m²) = (1/2) M (0.022 m)²

M = 3.05 kg

(a) Magnitude of the pulley's angular acceleration will be given as :

using a formula, we have

$\Delta$$\theta$ = $\omega$₀ t + (1/2) $\alpha$ t²

(1 rad) = (0) t + (0.5) $\alpha$ (0.179 s)²

$\alpha$ = (1 rad) / (0.0160205 s²)

$\alpha$ = 62.4 rad/s²

(b) Magnitude of either block's acceleration will be given as :

we know that, a = R $\alpha$ = (0.022 m) (62.4 rad/s²)

a = 1.37 m/s²

(c) The string tension, T₁ which is given as :

For Box 1 (hanging), we have

m₁ g - T₁ = m₁ a                                                                    { eq.1 }

T₁ = m₁ (g - a) = (5.6 kg) [(9.8 m/s²) - (1.37 m/s²)]

T₁ = 47.2 N

(d) The string tension, T₂ which is given as :

For Box 2 (table surface), we have

T₂ - F_f = m₂ a                                                               { eq.2 }

For a pulley, we have

R T₁ - R T₂ = I $\alpha$

R (T₁ - T₂) = (1/2) M R² (a / R)

(T₁ - T₂) = (1/2) M a                                           

T₂ = T₁ - (1/2) M a                                                        { eq.3 }

T₂ = (47.2 N) - (0.5) (3.05 kg) (1.37 m/s²)

T₂ = (47.2 N) - (2.08925 N)

T₂ = 45.1 N