Question 21.3: A 5-N · m torque is applied to the vertical shaft CD shown i......

The principle of work and energy may be used for the solution. Why?

Work. If shaft CD, the axle CE, and gear A are considered as a system of connected bodies, only the applied torque M does work. For two revolutions of CD, this work is ∑U_{1-2} = (5 N· m)( 4π rad) = 62.83 J.

Kinetic Energy. Since the gear is initially at rest, its initial kinetic energy is zero. A kinematic diagram for the gear is shown in Fig. 21-10b. If the angular velocity of CD is taken as ω_{CD}, then the angular velocity of gear A is ω_{A} = ω_{CD} + ω_{CE}. The gear may be imagined as a portion of a massless extended body which is rotating about the fixed point C. The instantaneous axis of rotation for this body is along line CH, because both points C and H on the body (gear) have zero velocity and must therefore lie on this axis. This requires that the components ω_{CD} and ω_{CE} be related by the equation ω_{CD}/0.1 m =  ω_{CE} /0.3 m or  ω_{CE} = 3ω_{CD}. Thus,

ω_{A} =  – ω_{CE}i  + ω_{CD}k  = – 3ω_{CE}i  +  ω_{CD}k      (1)

The x, y, z axes in Fig. 21-10a represent principal axes of inertia at C
for the gear. Since point C is a fixed point of rotation, Eq. 21-1S may
be applied to determine the kinetic energy, i.e.,

T = \frac{1}{2}  I_{x}ω^2_{x}   +  \frac{1}{2}  I_{y}ω^2_{y} + \frac{1}{2}  I_{z}ω^2_{z}                     (2)

Using the parallel-axis theorem, the moments of inertia of the gear about point C are as follows:

I_{x} = \frac{1}{2}  (10 kg)(0.1 m)² = 0.05   kg· m²

I_{y} = I_{z} = \frac{1}{4}(10    kg)(0.1   m)² + 10kg(0.3   m)² = 0.928   kg· m² 

Since ω_{x} = -3ω_{CD}, ω_{y} =  0, ω_{z} = ω_{CD}, Eq. 2 becomes

T_{A} = \frac{1}{2} (0.08)( -3ω_{CD})²  +  0  +   \frac{1}{2}(0.925)(ω_{CD})² = 0.6875ω²_{CD}

Principle of Work and Energy. Applying the principle of work and energy, we obtain

T_{1} + ∑U_{1-2} = T_{2}

o  +   62.83 = 0.6875ω²_{CD}

ω_{CD} = 9.56 rad/s