Question

A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.5 cm and mass 0.79 kg .
A. If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
B. In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s , it comes to rest in 6.4 s . What is the average frictional torque acting on the pulley?

Answer 1

Answer:

Explanation:

Let the acceleration be a of the system

T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass

for motion of 3.85 kg , applying newton's law

3.85g - T₁ = 3.85 a

for motion of 2.6 kg

T₂ - 2.6g = 2.6 a

T₂ - T₁ + 1.25 g = 6.45 a

T₁ - T₂ = 1.25 g -  6.45 a

for motion of pulley

(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration

(T₁ - T₂ ) x R = 1 /2  m R² x a / R

(T₁ - T₂ )  =   m  x a / 2 = .79 x a / 2 = . 395 a

1.25 g -  6.45 a = .395 a

1.25 g = 6.845 a

a = 1.79 m /s²

B )

When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s

Average deceleration = .2 / 6.4 = .03125 m /s²

Total deceleration created by frictional torque = 1.79 + .03125

= 1.82125 m /s²

If R be the average frictional torque  acting on the pulley

angular deceleration of pulley = a / R

= 1.82125 / .045

= 40.47 rad /s²

Now  R = I x 40.47 , I is moment of inertia of pulley

= 1 /2 x .79 x .045² x 40.47

= .0323 N.m

Torque created = .0323 Nm

Answer 2

The acceleration of the two masses hanging from ends of the pulley is 31 m/s².

The average frictional torque acting on the pulley is 0.55 Nm.

The given parameters;

• mass of the pulley, = M = 0.79 kg
• first mass, m₁ = 3.85 kg
• second mass, m₂ = 2.6 kg
• radius, R = 4.5 cm = 0.045 m

The acceleration of the two masses is determined by taking net torque acting on the pulley;

$\tau _{net} = I \alpha$

$T_1R - T_2R = I \alpha$

where;

• T is the tension on both stings suspending the masses = mg
• I is the moment of inertia of the pulley $= \frac{MR^2}{2}$
• α is the angular acceleration

$R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)$

Substitute the given parameters, to solve for the acceleration of the masses;

$a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2$

The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.

$a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm$

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