A homogeneous disk of radius r and mass m is mouted on an axle OG of length L and negligible mass. The axle is pivoted at the fixed point O, and the disk is constrained to roll on a horizontal floor. knowing that the ‘disk’ rotates counterclockwise at the rate $ω_{1}$ about the axle OG,

Choose the correct option at the given 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

The X component of angular momentum of disc about ‘O’ is $\frac{1}{2} m r^{2} ω_{1}$

The Z component of angular momentum of disc about ‘O’ is zero

The angular velocity of the disk $ω_{i} \hat{i} - \frac{ω_{1} r}{L} \hat{j}$

The angular velocity of the disk $- (\frac{ω_{1} r}{L}) \hat{j}$

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$ω = ω_{1} i - ω_{2} j$
Question Image
$\begin{array}{l} V_{C} {=ω×r}_{c} =0 \\ (ω_{1} {i-ω}_{2} j) × (Li-rj) =0 \\ ({Lω}_{2} {=rω}_{1}) {k=0 ω}_{2} {=ω}_{1} i/L \\ {ω=ω}_{1} i- ({rω}_{1} /L) j \\ H_{X} {=I}_{X} ω_{X} = (\frac{1}{2} {mr}^{2}) ω_{1} \\ H_{Z} {=I}_{Z} ω_{Z} = ({mL}^{2} + \frac{1}{4} {mr}^{2}) 0=0 \end{array}$