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Anastaziya [24]
3 years ago
13

A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation

matrix that accomplishes these rotations in the given order.
Engineering
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

R = \left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

Let \theta and \alpha be the the angles 60⁰ and 30⁰ respectively

that is \theta = 60⁰ and

\alpha = 30⁰

The matrix is given by the following expression:

\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

The angles can be evaluated and left in the surd form.

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To solve this we use the energy balance equation, that is:

E_{in}-E_{out}=\Delta E_{sys}\\\\Q_{in}+W_{in}+W_{out}=\Delta U\\\\Q_{in}+W_{in}=\Delta U\\\\But\ W_{in}=Voltage(V)*Current(I)*change\ in\ time(\Delta t)=VI\Delta t,\Delta U=m(h_2-h_1)\\\\Given\ that\ m=15kg,V=110\ V,\Delta t=6\ min = (6*60\ s)=360\ s,Q_{in}=20.67\ kJ/kg*15\ kg=310.05\ kJ=310050\ J\\\\From\ table: At\ P_1=280kPa, h_1=249.71\ kJ/kg=249710\ J/kg;At\ P_2=280kPa \ and\  T_1=75^oC,P_2=319.95\ kJ/kg=319950\ J/kg\\\\Substituting:\\\\310500+(110*360*I)=15(319950-249710)\\\\

39600I=1053600-310500\\\\39600I=743100\\\\I=18.77\ A

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