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Byju's Answer

Standard VIII

Mathematics

Application of Negative Numbers in Speed Distance Time

A rod of leng...

Question

A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60 degrees with the horizontal plane, the velocity of point A on the rod is $1 \frac{m}{s}$ . The angular velocity of the rod at the instant is

2 \frac{r a d}{s}

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1.5 \frac{r a d}{s}

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0.5 \frac{r a d}{s}

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0.75 \frac{r a d}{s}

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Solution

The correct option is A $2 \frac{r a d}{s}$

The instantaneous center of AB will be perpendicular to $V_{A}$ and $V_{B}$ .
$I A = 1 cos θ = 1 cos 60^{\circ} = \frac{1}{2} m$

$V_{A} = ω \times I A = ω \times r_{A}$

or $ω = \frac{V_{A}}{r_{A}} = \frac{1}{1 / 2} = 2 r a d / s$

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A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60 degrees with the horizontal plane, the velocity of point A on the rod is 1ms. The angular velocity of the rod at the instant is

The correct option is A 2rads The instantaneous center of AB will be perpendicular to VA and VB. IA=1cosθ=1cos60∘=12m VA=ω×IA=ω×rA or ω=VArA=11/2=2rad/s

A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60 degrees with the horizontal plane, the velocity of point A on the rod is $1 \frac{m}{s}$ . The angular velocity of the rod at the instant is

The correct option is A $2 \frac{r a d}{s}$

The instantaneous center of AB will be perpendicular to $V_{A}$ and $V_{B}$ .
$I A = 1 cos θ = 1 cos 60^{\circ} = \frac{1}{2} m$

$V_{A} = ω \times I A = ω \times r_{A}$

or $ω = \frac{V_{A}}{r_{A}} = \frac{1}{1 / 2} = 2 r a d / s$