The instantaneous angular position of a point on a rotating wheel is given by the equation $θ (t) = 2 t^{3} - 6 t^{2}$ . Find the angular acceleration for t = 1 sec.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

0 rad/sec²

12 rad/sec²

1 rad/sec²

6 rad/sec²

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given:

$\begin{array}{l} θ = 2 t^{3} - 6 t^{2} \\ \Rightarrow ω = \frac{d θ}{d t} = 6 t^{2} - 12 t \\ \Rightarrow α = \frac{d ω}{d t} = 12 t - 12 \end{array}$
For t = 1 sec,

$\begin{array}{l} \Rightarrow α = 12 - 12 \end{array} = 0 rad / s^{2}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!