A tappet valve mechanism in an IC engine comprises a rocker arm ABC that is hinged at B as shown in the figure. The rocker is assumed rigid and it oscillates about the hinge B. The mass moment of inertia of the rocker about B is 10-4 kg.m2. The rocker arm dimensions are a = 3.5 cm and b = 2.5 cm. A pushrod pushes the rocker at location A, when moved vertically by a cam that rotates at N rpm. The pushrod is assumed massless and has a stiffness of 15 N/mm. At the other end C, the rocker pushes a valve against a spring of stiffness 10 N/mm. The valve is assumed massless and rigid.

Resonance in the rocker system occurs when the cam shaft runs at a speed of ______ rpm (round off to the nearest integer).

The correct option is 4739.

1. Identify the given parameters from the problem:
Mass moment of inertia of the rocker arm about B:
$$I_B={10}^{-4}{\text{kg·m}}^2$$
Distance from hinge B to pushrod A:
$$a=3.5\text{cm}=0.035\text{m}$$
Distance from hinge B to valve C:
$$b=2.5\text{cm}=0.025\text{m}$$
Stiffness of the pushrod at A:
$$k_1=15\text{N/mm}=15,000\text{N/m}$$
Stiffness of the valve spring at C:
$$k_2=10\text{N/mm}=10,000\text{N/m}$$

2. Formulate the Equation of Motion:
Let us assume the rigid rocker arm ABC undergoes a small angular displacement $\theta$ about the hinge B.
- The vertical displacement at point A is $x_A=a\theta$, resulting in a restoring torque of $T_1=-k_1{a}^2\theta$ about B.
- The vertical displacement at point C is $x_C=b\theta$, resulting in a restoring torque of $T_2=-k_2{b}^2\theta$ about B.
Applying D'Alembert's principle or writing the torque balance equation:
$$I_B\ddot{\theta }+\left(k_1{a}^2+k_2\right)\theta =0$$

3. Calculate the Equivalent Torsional Stiffness:
The equivalent torsional stiffness $k_t$ is given by:
$$k_t=k_1{a}^2+k_2{b}^2$$
Substitute the numerical values:
$$k_t=\left(15,000\right)\times {\left(0.035\right)}^2+\left(10,000\right)\times {\left(0.025\right)}^2$$
$$k_t=15,000\times 0.001225+10,000\times 0.000625$$
$$k_t=18.375+6.25=24.625\text{N·m/rad}$$

4. Calculate the Natural Frequency:
The angular natural frequency ${\omega }_n$ of the system is:
$${\omega }_n=\sqrt{\frac{k_t}{I_B}}=\sqrt{\frac{24.625}{{10}^{-4}}}=\sqrt{246,250}\approx 496.236\text{rad/s}$$

5. Calculate the Cam Shaft Speed for Resonance:
Resonance in the system occurs when the excitation frequency from the cam rotation is equal to the natural frequency.
Converting the natural frequency to revolutions per minute (rpm):
$$N=\frac{60\times {\omega }_n}{2\pi }$$
$$N=\frac{60\times 496.236}{2\times 3.14159}\approx 4738.7\text{rpm}$$
Rounding off to the nearest integer, we get:
$$N\approx 4739\text{rpm}$$