Question

8. Mass of a rod of length 4 m is 3 kg. Its centre of gravity
is 1.5 m from one of its ends. At the other end a mass of 8kg is tied up. the rod is at horizontal position . it is turned to a vertical position in such a way that the 8kg mass goes to the top . what is the work done in changing the orientation of the rod ?
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Answer 1

Answer:

Let's consider that the rod and the mass are horizontal initially, so the whole system is at y=0 initially.(This can be assumed as nothing specific has been given about the length of the string with which the 8 kg block has been tied to one of the end of the rod, so one can with no problem consider that the 8 kg block is just like attached to that other end of the rod)

Now, to make this rod vertical upwards so that the 8 kg block is at the top, in the final configuration

The height of center of gravity of the rod would have been changed by 1.5 metres from y=0 initial horizontal reference line and because the block is at the top, it's center of gravity would have changed its height by 4 metres.

So, Finally,

Work done by external agent = Change in potential energy of the block rod system

= m1gh1 + m2gh2

= (3)(10)(1.5)+(8)(10)(4)

= 45 + 320 = 365 Joules

I've considered g = 10 m/s^2 for calculation to be easier, you can put in g = 9.8 m/s^2 or (9.81 m/s^2 to be more specific) you'll get and match the exact value of the answer that has been given

Hope this helps you !

If you've have any doubt in the steps or understanding of the sequence of steps followed here, you can ask a question related to that on your profile, other users as well as me will keep a track and answer if possible ! ^_^

Answer 2

Solution 1:-

We know the work done in rotating a system from an inclination $\sf{\theta_1}$ to an inclination $\sf{\theta_2}$ is,

$\displaystyle\mathsf{\longrightarrow W=\int\limits_{\theta_1}^{\theta_2}}\vec{\sf{\tau}}\cdot\vec{\sf{d\theta}}$

where $\vec{\sf{\tau}}$ is the net torque acting on the system.

In case of rotating a system $\vec{\sf{\tau}}$ and $\vec{\sf{d\theta}}$ are in same direction.

Thus,

$\displaystyle\mathsf{\longrightarrow W=\int\limits_{\theta_1}^{\theta_2}}\sf{\tau\ d\theta}$

Since $\displaystyle\sf{\tau=\sum rF\sin\theta,}$

$\displaystyle\mathsf{\longrightarrow W=\int\limits_{\theta_1}^{\theta_2}}\sf{\sum rF\sin\theta\ d\theta}$

$\displaystyle\mathsf{\longrightarrow W=\sum rF\int\limits_{\theta_1}^{\theta_2}}\sf{\sin\theta\ d\theta}$

$\displaystyle\mathsf{\longrightarrow W=-\big[\cos\theta\big]_{\theta_1}^{\theta_2}\sum rF}$

$\displaystyle\mathsf{\longrightarrow W=\left(\cos\theta_1-\cos\theta_2\right)\sum rF}$

In the question the system is rotated from $\sf{\theta_1=0^o}$ to $\sf{\theta_2=90^o.}$

And,

$\displaystyle\sf{\longrightarrow \sum rF=1.5\times3\times9.8+4\times8\times9.8}$

$\displaystyle\sf{\longrightarrow \sum rF=357.7\ N\,m}$

Then,

$\displaystyle\mathsf{\longrightarrow W=\left(\cos0^o-\cos90^o\right)357.7\ J}$

$\displaystyle\mathsf{\longrightarrow W=\left(1-0\right)357.7\ J}$

$\sf{\longrightarrow\underline{\underline{W=357.7\ J}}}$

Solution 2:-

The system was initially arranged in horizontal position.

The center of mass of the rod will be at zero height from the bottom.

So is the 8 kg mass tied at one end of the rod.

So, initial potential energy of the system would be,

$\sf{\longrightarrow U_i=3\times9.8\times0+8\times9.8\times0}$

$\sf{\longrightarrow U_i=0\ J}$

After the system being arranged in vertical position, the center of mass of the rod is at a height 1.5 m from bottom.

And the 8 kg mass is at 4 m from the bottom.

So final potential energy of the system,

$\sf{\longrightarrow U_f=3\times9.8\times1.5+8\times9.8\times4}$

$\sf{\longrightarrow U_f=357.7\ J}$

The work done in changing the orientation of the system results in this change in potential energy.

Therefore,

$\sf{\longrightarrow W=U_f-U_i}$

$\sf{\longrightarrow W=357.7\ J-0\ J}$

$\sf{\longrightarrow\underline{\underline{W=357.7\ J}}}$

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