Dark mode: OFF

Comments

**Given data:**

Moment of inertia, $\ I = 25 \ kg \sdot m^2$

Stiffness of torsional spring $= 20 \ N \sdot m/rad$

$C_c = (?)$

According to D'Alembert's principle;

$\Sigma \bigg[ \ inertia \ torque + \ external \ torque \bigg] = 0$ $\begin{aligned} \therefore \ I \sdot \ddot \theta + (c_c \dot x) \sdot a + K_t \theta &= 0\\ \therefore \ I \sdot \ddot \theta + c_ca \dot \theta a + K_t \theta &= 0 \quad (x = a \theta \ and \ \dot x = a \dot \theta)\\ \therefore \ I \sdot \ddot \theta + c_ca^2 \dot \theta + K_t \theta &= 0 \end{aligned}$The above equation can be written as:

$\therefore \ I \sdot \ddot \theta \ + c_t \ddot \theta \ + K_t \theta = 0 \quad \quad \text {where} \ \ c_t = c_ca^2$Now,

$\begin{aligned} c_t &= 2 \sdot I \sdot \omega_n\\ \therefore \ c_t &= 2 \sdot \ I \sqrt{K_t \over I}\\ \therefore \ c_c \sdot a^2 &= 2 \sqrt{K_tI} \quad \because (c_t = c_ca^2)\\ \therefore \ c_c &= {2 \over a^2} \sqrt{K_tI}\\ &= {2 \over (0.05)^2} \sqrt{20 \times 25}\\ \therefore \ c_c &= 17888.54 \ N \sdot s/m \end{aligned}$If our notes and videos are helpful to you, kindly support us by making a donation from our support page so we can continue making more content for students like you.

Go to support page

- Balancing of V-Engines
- Damped free Vibration - Numerical 1
- Critically Damped System (ξ = 1)
- Logarithmic Decrement (δ)
- Over-Damped System (ξ>1)
- Under-Damped System (ξ < 1)
- Static and Dynamic Balancing
- Concept of Direct and Reverse Crank for V-engines & Radial engines
- Numerical - Multi Cylinder Inline Engine (with firing order)

Sign in with google to add a comment

Sign in to post a commentBy signing in you agree to Privacy Policy

All comments that you add will await moderation. We'll publish all comments that are topic related, and adhere to our Code of Conduct.

Want to tell us something privately? Contact Us

No comments available.

## Comments: