Respuesta :
Answer:
a)Time taken will be 2.783 s
b)Time taken will be 2.564 s
Explanation:
a)Since the pulley has mass ,
- It will have a MOMENT OF INERTIA . in other terms, whn the string slides upon it, it will produce a torque ( due to the tension) and thus it will make the pulley roll.
- The string soesn't get slackened. Thus the acceleration along the string must be constant - which is the string constraint.
- the FBD's of the bodies are attached. from them ,
[tex]m_{1}g-T_{1}= m_{1}a//T_{2}-m_{2}g=m_{2}a[/tex] ------3
- Since the string doesn't slip, the acceleration of the pulley at the end point of contact of the string must be equal to [tex]a[/tex]
- or,
αR = a ; ------1
- writing the torque equation about COM of the pulley , we get
[tex](T_{1} - T_{2})*R=m*R^{2}*(alpha)=5*R^{2}*\frac{a}{R}[/tex] ------2
solving these we get ,
[tex]a=\frac{4g}{33}[/tex]
∴
a)time taken :
[tex]s=ut+\frac{1}{2} at^{2}\\u=0\\a=\frac{4g}{33} \\s=4.6\\4.6=\frac{1}{2} * \frac{4*9.8}{33} *t^{2}\\t= 2.783 sec[/tex]
ANS : 2.783 sec
b)
In case of B, mass is zero.
Thus, there is no rotation of the pulley. this is equivalent to a normal 1 Dimension motion question
- equations are:
[tex]m_{1}g-T_{1}=m_{1}a\\T_{1}-m_{2}g=m_{2}a\\a= \frac{m_{1}-m_{2}}{m_{1}+m_{2}}g\\a=\frac{4g}{28} \\a=\frac{g}{7} \\a=1.4ms^{-2}[/tex]
Thus time t will be ,
[tex]s=\frac{1}{2} at^{2}\\4.6=\frac{1}{2}*1.4*t^{2}\\ t=2.564 sec[/tex]
ANS : 2.564 sec
The time it takes the block to reach the ground can be found by making
use of the law of conservation of energy.
- a. The time it takes the the block m₁ to hit the floor is approximately 2.596 seconds.
- b. The time it takes the the block m₁ to hit the floor, if the mass of the pulley were neglected is approximately 2.562 seconds.
Reasons:
Given parameters are;
Mass of block m₁ = 16.0 kg
Mass of block m₂ = 12.0 kg
Mass of the pulley, M = 5.00 kg
By conservation of energy, we have;
m₁g·h - m₂·g·h = 0.5×(m₁ + m₂)·v² + 0.5·I·ω²
[tex]\omega = \dfrac{v}{R}[/tex]
[tex]Moment \ of \ inertia\ of \ pulley, I = \dfrac{1}{2} \cdot M \cdot R^2[/tex]
Therefore;
[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2 + 0.5 \cdot I \cdot \dfrac{v}{R}[/tex]
Which gives;
(16 - 12)×9.81×4.6 = 0.5×(16+12)×v² + 0.5×(0.5×5×0.3²)× [tex]\left(\dfrac{v}{0.3} \right)^2[/tex]
Solving gives, v ≈ 21.93 m/s
We have;
v ≈ 3.544
v² = 2·a·h
[tex]a = \dfrac{v^2}{2 \times h}[/tex]
Which gives;
[tex]a = \dfrac{3.544^2}{2 \times 4.6} \approx 1.365[/tex]
v = a×t
[tex]t = \dfrac{v}{a} = \dfrac{3.544}{1.365} \approx 2.596[/tex]
The time it takes the the block m₁ to hit the floor, t ≈ 2.596 seconds
b. When the mass of the pulley is neglected, we have;
[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2[/tex]
(16 - 12)×9.81×4.6 = 0.5×(16+12)×v²
180.504 = 14·v²
[tex]v = \sqrt{\dfrac{180.504}{14} } \approx 3.591[/tex]
[tex]a = \dfrac{3.591^2}{2 \times 4.6} \approx 1.401[/tex]
[tex]t = \dfrac{v}{a} = \dfrac{3.591}{1.401} \approx 2.562[/tex]
The time it takes the the block m₁ to hit the floor, if the mass of the pulley were neglected, t ≈ 2.562 seconds.
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