Question

Please explain what formulas to use and please dont skip steps so I can understand the process. Thank You

Problem 2: The 15.0 kg mass on a 40.0° frictionless inclined plane is tied to a string that is wrapped around a frictionless pulley (disk, I = MR?) of mass 12.0 kg and radius 9.00 cm. What is the angular acceleration of the disk after the 15.0 kg mass is released? al Ins 11 (F. ersona Questi blogy

Answer 1

Hi,

Hope you are doing well.

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FREE BODY DIAGRAM:

T sa mgsine mgcos mg

The weight of the block (mg) can be resolved into 2 perpendicular components (mgsin$\theta$and mgcos$\theta$) as shown in the free body diagram.

T= Tension on the string.

Let be the acceleration of the block downwards.

Given data:

Mass of the block, $m=15\;kg$

Acceleration due to gravity,

g9.8 m/s

Mass of the pulley, $M=12\;kg$

Radius of the pulley, $R=9\;cm=0.09\;m$

Angle of inclination, $\theta=40^{o}$

From free body diagram we have:

$ma=mg\sin\theta-T$

$\therefore T=mg\sin\theta-ma\rightarrow \boldsymbol{(1)}$

For the pulley:

Let $\alpha$ be the angular acceleration of the pulley.

We have, Angular acceleration x Radius = Tangential acceleration.

$\alpha R=a$

Also, Torque = Tangential force x Radius.

In this case, tangential force is the tension on the string.

$\therefore \tau=T.R\rightarrow \boldsymbol{(2)}$

Also, we have, $\tau=I\alpha\rightarrow \boldsymbol{(3)}$

Where, given that:

$I=\frac{1}{2}MR^{2}\rightarrow (4)$

From (2) and (3):

$I\alpha=T.R$

Substituting from (1) and (4):

$\frac{1}{2}MR^{2}\alpha=\left (mg\sin\theta-ma \right )R$

We have, $\alpha R=a$

$\therefore \frac{1}{2}MR\alpha=\left (mg\sin\theta-m\alpha R \right )$

$\frac{1}{2}MR\alpha+m\alpha R=mg\sin\theta$

Taking $\alpha$ as common:

$\alpha\left (\frac{1}{2}MR+m R \right )=mg\sin\theta$

Substituting from given values:

$\alpha\left (\frac{1}{2}\times 12 \;kg\times0.09\;m+15\;kg\times0.09\;m \right )=15\;kg\times9.8\;m/s^{2}\sin40^{o}$

$\alpha=\frac{15\;kg\times9.8\;m/s^{2}\sin40^{o}}{1.89\;kg.m}$

$\boldsymbol{\therefore \alpha=49.99\;rad/s^{2}\approx50\;rad/s^{2}}$

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Hope this helped for your studies. Keep learning. Have a good day.

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