A hoop is placed on the rough surface such that it has an angular velocity $ω = 4 rad / s$ and an angular deceleration $α = 5 rad / s^{2}$ . Also, its centre has a velocity of $v_{0} = 5 m / s$ and a deceleration $a_{0} = 2 m / s^{2}$ . Determine the magnitude of acceleration of point B at this instant.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

(1) $8.21 m / s^{2}$

(2) $4.21 m / s^{2}$

(3) $6.21 m / s^{2}$

(4) $2.21 m / s^{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$a_{B} = a_{0} + a_{B / 0}$

Here, $a_{B / 0}$ has two components $a_{r}$ (tangential acceleration) and $a_{n}$ (normal acceleration)

$a_{t} = rα = (0.3) (5) = 1.5 m / s^{2} a_{n} = {rω}^{2} = (0.3) (4)^{2} = 4.8 m / s^{2} and a_{0} = 2 m / s^{2} ∴ a_{B} = \sqrt{{({Σa}_{x})}^{2} + {({Σa}_{y})}^{2}} = \sqrt{{(2 + 4.8 \cos 45^{0} - 1.5 \sin 45^{0})}^{2} + {(4.8 \sin 45^{0} + 1.5 \sin 45^{0})}^{2}} = 6.21 m / s^{2}$