Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

A pulley fixed to the ceiling of an elevator car carries a thread whose ends are attached to the loads of masses

m_{1}

and

m_{2}

. The car starts going up with an acceleration

\vec{w_{0}}

. Assuming the masses of the pulley and the thread, as well as the friction, to be negligible find,

(a)

the acceleration of the load

m_{1}

relative to the elevator shaft and relative to the car,

(b)

the force exerted by the pulley on the ceiling of the car.

Accepted Answer

Let us assume upward direction as positive direction and downward direction as negative direction.

Given,
Acceleration of pulley(or lift) $= w_{0}$
Mass of blocks inside the pulley $= m_{1} and m_{2}$

Suppose w.r.t. ground, acceleration of mass $m_{1}$ is $w_{1}$ in downward direction. Then acceleration of mass $m_{2}$ will be $w_{2}$ in the upward direction. Acceleration of blocks w.r.t. pulley is $w_{\frac{1}{P}} and w_{\frac{2}{P}}$ respectively.
W.r.t. lift, movement of string will be same for both sides of the pulley.Using string constraints-
${\vec{w}}_{\frac{1}{P}} = - {\vec{w}}_{\frac{2}{P}} \dots (1)$ (relating acceleration of blocks w.r.t. pulley)

Now using equations of relative motion for blocks $m_{1}$ and $m_{2} -$
${\vec{w}}_{1} = {\vec{w}}_{0} + {\vec{w}}_{\frac{1}{P}} \dots (2)$
${\vec{w}}_{2} = {\vec{w}}_{0} + {\vec{w}}_{\frac{2}{P}} \dots (3)$

Using equation $(1), (2) and (3) -$

$\Rightarrow {\vec{w}}_{1} + {\vec{w}}_{2} = 2 {\vec{w}}_{0}$

Putting value of ${\vec{w}}_{1} and {\vec{w}}_{2}$ with sign-
$\Rightarrow w_{0} = \frac{w_{2} - w_{1}}{2} \dots (4)$

Using Newton's second law for both the blocks in the ground frame

$m_{1} g - T = m_{1} w_{1} \dots (5)$

$T - m_{2} g = m_{2} w_{2} \dots (6)$

Using equations $(4), (5) and (6) -$
$w_{1} = \frac{(m_{1} - m_{2}) g + 2 m_{2} w_{0}}{m_{1} + m_{2}}$ and $T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} (g + w_{0})$

This is acceleration of block $m_{1}$ w.r.t. ground. Acceleration of Block $m_{1}$ w.r.t. pulley is given by-
${\vec{w}}_{\frac{1}{P}} = {\vec{w}}_{\frac{1}{g r o u n d}} + {\vec{w}}_{\frac{g r o u n d}{P}} = {\vec{w}}_{1} - {\vec{w}}_{P}$
$\Rightarrow {\vec{w}}_{\frac{1}{P}} = - \{\frac{(m_{1} - m_{2}) g + 2 m_{2} w_{0}}{m_{1} + m_{2}}\} - (w_{0})$
$\Rightarrow w_{\frac{1}{P}} = - \{(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) (g + w_{0})\}$

Negative sign shows that acceleration will be in the downward direction.

$(b)$ From the diagram, we can see that the force applied by the pulley on the lift is $2 T$ in the downward direction,

$F = 2 T = \frac{4 m_{1} m_{2}}{m_{1} + m_{2}} (g + w_{0})$ .

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