A piston is connected by a rod of #14 cm# to a crankshaft at a point #5 cm# away from the axis of rotation. Determine how fast the crankshaft is rotating when the piston is 11 cm away from the axis of rotation and is moving toward it at 1200 cm/s?

Question

Aurora Alvarado · Accepted Answer

#(d theta)/dt = 325.114 color(white)"." "rad"/s#

This problem is asking us to find #(d theta)/dt#, or the angular velocity of the piston.

Let #Q = overline(OQ)#

So we know that:

#(dQ)/dt]_(Q= 11) = -1200"cm"/s#

Before we start, let's write out a formula for theta given what we know, using the Law of Cosines:

#14^2 = 5^2 + Q^2 - 2(5)(Q)cos(theta)#

#10Qcostheta = Q^2 - 171#

#10costheta = Q - 171/Q#

Now, take the derivative of this with respect to time:

#-10sintheta * (d theta)/dt = (dQ)/dt + 171/Q^2 * (dQ)/dt#

Since we already have values for #Q# and #(dQ)/dt#, all we need to do is find #sintheta# when #Q = 11#.

To do that, let's use the Law of Cosines again, but manipulate it further until we get #sintheta#. (This time, we know #Q = 11#)

#14^2 = 5^2 + 11^2 - 2(5)(11)cos(theta)#

#50 = -110cos(theta)#

#-5/11 = cos(theta) = sqrt(1-sin^2theta)#

#25/121 = 1-sin^2theta#

#96/121 = sin^2theta#

#(4sqrt6)/11 = sintheta#

Now we know that when #Q=11#, #(dQ)/dt = -1200# and #sintheta = (4sqrt6)/11#.

This is all we need to solve for #(d theta)/dt#:

#-10sintheta * (d theta)/dt = (dQ)/dt + 171/Q^2 * (dQ)/dt#

#-10((4sqrt6)/11)*(d theta)/dt = -1200 + 171/121(-1200)#

#-(40sqrt6)/11*(d theta)/dt = -1200(292/121)#

#(d theta)/dt = (1200 * 292 * 11)/(40sqrt6 * 121)" ""rad"/s#

#(d theta)/dt = (8760)/(11sqrt6) = (8760sqrt6)/66 = (1460sqrt6)/11 " ""rad"/s#

Or if you prefer decimal form:

#(d theta)/dt = 325.114 " " "rad"/s#

Emilia Aguilar · Answer

undefined To determine the angular velocity of the crankshaft, use the formula: ω = v / r, where ω is the angular velocity, v is the velocity of the piston, and r is the distance from the piston to the axis of rotation. Given v = 1200 cm/s and r = 11 cm, plug in the values to find ω.