LOG#042. Pulley with variable mass.

This interesting problem was recently found in certain public examination. I will solve it here since I found it fascinating and useful, since it is a problem with “variable mass”.

A pulley with negligible mass is given (see the figures above). Its radius is R. We tie two masses, m_1 and m_2 to the extremes of the pulley with the aid of a string. The string has a linear mass density \lambda and it has a total length equal to L. Calculate:

a) The forces in the extremes of the pulley over the masses and the acceleration when we release the system from the rest.

b) The velocity of the masses and the position of a the masses as a function of time.

Remark: We have the contraint q_2+\pi R+q_1=L. L is the total length of the string, R is the radius of the pulley, and q_2 and q_1 are the pieces of string unwrapped to the pulley for the masses m_2 and m_1, respectively. We can suppose that initially q_1=0 and q_2=l. Note that due to this constraint \dot{q_1}+\dot{q_2}=0, where the dot denotes derivative with respect to time, i.e., velocity of both masses, respecively too.

Solution.

a) Suppose m_1>m_2. Thus, the forces acting on the masses are:

\boxed{F_1=m_1 g+\lambda gx-T=\overline{m_1}a=(m_1+\lambda x)a}

\boxed{F_2=T-m_2 g-\lambda g (l-x)=\overline{m_2}a=(m_2+(L-x)\lambda) a}

Adding the two equations, we get

m_1g-m_2g+2\lambda g x-\lambda g l=(m_1+m_2+L\lambda)a

and then the acceleration will be

\boxed{a=\dfrac{m_1g-m_2g-\lambda gl-2\lambda gx}{m_1+m_2+\lambda L}=\dfrac{(m_1-m_2)g-l\lambda g+2g\lambda x}{m_1+m_2+\lambda L}}

b) Firstly, we have to derive the differential equations for the system with variable mass.
The equations for both masses are, respectively, using Newton’s laws:

(m_1+\lambda x)x''=m_1 g+\lambda gx-T

(m_2+(L-x)\lambda)x''=T-m_2g-\lambda g(L-x)

Adding these two equations we get the differential equation to be solved:

\boxed{a=x''=\dfrac{m_1g-m_2g-\lambda g L+2g\lambda x}{m_1+m_2+\lambda L}}

with x(0)=x'(0)=0

We can solve the differential equation

x''=\dfrac{(m_1-m_2)g-L\lambda g+2g\lambda x}{m_1+m_2+\lambda L}

with the initial conditions v(0)=x'(0)=0 and x(0)=0 easily, to obtain

\boxed{v(t)=x'(t)=\dfrac{((m_1-m_2)-\lambda L)\sqrt{\lambda g}}{\lambda \sqrt{2(m_1+m_2+\lambda L)}}\sinh \left(\sqrt{\dfrac{2\lambda g}{m_1+m_2+\lambda L}}t\right)}

\boxed{x(t)=\dfrac{(m_1-m_2-\lambda L)}{\lambda}\sinh^2 \left(\sqrt{\dfrac{\lambda g}{2(m_1+m_2+\lambda L)}}t\right)}

As the mass of the string depends on the linear density, the usual procedure to obtain the variation of the kinetic energy as the work done by the potential forces is something more complicated:

\Delta E_c=\Delta W (x)=\int_0^x F(x)dx

It can be shown that

\dfrac{1}{2}(m_1+m_2+\lambda L)v^2=(m_1-m_2+\lambda x-\lambda l)g x

v=\sqrt{\dfrac{2gx(m_1-m_2+\lambda x-\lambda l)}{m_1+m_2+\lambda L}}

Please, note that in this case, we have to be very careful with l and L. They are related with a contraint, but they are not the same thing.

Advertisements


Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s