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Engineering Mechanics → Dynamics → Kinematics of Rigid Bodies
Last updated 5 September 2026
Question In a rigid body in plane motion, the point R is accelerating with respect to point P at
10 ∠ 180 ∘ m/s 2 10 \angle 180^\circ \text{ m/s}^2 10∠18 0 ∘ m/s 2 . If the instantaneous acceleration of point Q is zero, the acceleration (in
m/s 2 \text{m/s}^2 m/s 2 ) of point R is
Solution Let Q be the origin
( 0 , 0 ) (0,0) ( 0 , 0 ) . Then
P = ( 0 , 12 ) P = (0, 12) P = ( 0 , 12 ) and
R = ( 16 , 0 ) R = (16, 0) R = ( 16 , 0 ) . Given
a ⃗ Q = 0 \vec{a}_Q = 0 a Q = 0 and
a ⃗ R / P = 10 ∠ 180 ∘ = − 10 i ^ \vec{a}_{R/P} = 10 \angle 180^\circ = -10\hat{i} a R / P = 10∠18 0 ∘ = − 10 i ^ .
For a rigid body,
a ⃗ R = a ⃗ Q + α ⃗ × r ⃗ R / Q − ω 2 r ⃗ R / Q \vec{a}_R = \vec{a}_Q + \vec{\alpha} \times \vec{r}_{R/Q} - \omega^2 \vec{r}_{R/Q} a R = a Q + α × r R / Q − ω 2 r R / Q and
a ⃗ P = a ⃗ Q + α ⃗ × r ⃗ P / Q − ω 2 r ⃗ P / Q \vec{a}_P = \vec{a}_Q + \vec{\alpha} \times \vec{r}_{P/Q} - \omega^2 \vec{r}_{P/Q} a P = a Q + α × r P / Q − ω 2 r P / Q .
Subtracting gives
a ⃗ R / P = α ⃗ × r ⃗ R / P − ω 2 r ⃗ R / P \vec{a}_{R/P} = \vec{\alpha} \times \vec{r}_{R/P} - \omega^2 \vec{r}_{R/P} a R / P = α × r R / P − ω 2 r R / P .
With
r ⃗ R / P = 16 i ^ − 12 j ^ \vec{r}_{R/P} = 16\hat{i} - 12\hat{j} r R / P = 16 i ^ − 12 j ^ and
α ⃗ = α k ^ \vec{\alpha} = \alpha\hat{k} α = α k ^ :
− 10 i ^ = α k ^ × ( 16 i ^ − 12 j ^ ) − ω 2 ( 16 i ^ − 12 j ^ ) = ( 12 α − 16 ω 2 ) i ^ + ( 16 α + 12 ω 2 ) j ^ -10\hat{i} = \alpha\hat{k} \times (16\hat{i} - 12\hat{j}) - \omega^2 (16\hat{i} - 12\hat{j}) = (12\alpha - 16\omega^2)\hat{i} + (16\alpha + 12\omega^2)\hat{j} − 10 i ^ = α k ^ × ( 16 i ^ − 12 j ^ ) − ω 2 ( 16 i ^ − 12 j ^ ) = ( 12 α − 16 ω 2 ) i ^ + ( 16 α + 12 ω 2 ) j ^ .
Equating components:
1)
16 α + 12 ω 2 = 0 ⇒ ω 2 = − 4 3 α 16\alpha + 12\omega^2 = 0 \Rightarrow \omega^2 = -\frac{4}{3}\alpha 16 α + 12 ω 2 = 0 ⇒ ω 2 = − 3 4 α 2)
12 α − 16 ω 2 = − 10 ⇒ 12 α − 16 ( − 4 3 α ) = − 10 ⇒ 100 3 α = − 10 ⇒ α = − 0.3 rad/s 2 12\alpha - 16\omega^2 = -10 \Rightarrow 12\alpha - 16(-\frac{4}{3}\alpha) = -10 \Rightarrow \frac{100}{3}\alpha = -10 \Rightarrow \alpha = -0.3 \text{ rad/s}^2 12 α − 16 ω 2 = − 10 ⇒ 12 α − 16 ( − 3 4 α ) = − 10 ⇒ 3 100 α = − 10 ⇒ α = − 0.3 rad/s 2 .
Then
ω 2 = − 4 3 ( − 0.3 ) = 0.4 rad/s 2 \omega^2 = -\frac{4}{3}(-0.3) = 0.4 \text{ rad/s}^2 ω 2 = − 3 4 ( − 0.3 ) = 0.4 rad/s 2 .
Now,
a ⃗ R = α ⃗ × r ⃗ R / Q − ω 2 r ⃗ R / Q = ( − 0.3 k ^ ) × ( 16 i ^ ) − ( 0.4 ) ( 16 i ^ ) = − 4.8 j ^ − 6.4 i ^ \vec{a}_R = \vec{\alpha} \times \vec{r}_{R/Q} - \omega^2 \vec{r}_{R/Q} = (-0.3\hat{k}) \times (16\hat{i}) - (0.4)(16\hat{i}) = -4.8\hat{j} - 6.4\hat{i} a R = α × r R / Q − ω 2 r R / Q = ( − 0.3 k ^ ) × ( 16 i ^ ) − ( 0.4 ) ( 16 i ^ ) = − 4.8 j ^ − 6.4 i ^ .
Magnitude
∣ a ⃗ R ∣ = ( − 6.4 ) 2 + ( − 4.8 ) 2 = 8 m/s 2 |\vec{a}_R| = \sqrt{(-6.4)^2 + (-4.8)^2} = 8 \text{ m/s}^2 ∣ a R ∣ = ( − 6.4 ) 2 + ( − 4.8 ) 2 = 8 m/s 2 .
Angle
θ = 180 ∘ + tan − 1 ( 4.8 6.4 ) = 180 ∘ + 36.87 ∘ ≈ 217 ∘ \theta = 180^\circ + \tan^{-1}(\frac{4.8}{6.4}) = 180^\circ + 36.87^\circ \approx 217^\circ θ = 18 0 ∘ + tan − 1 ( 6.4 4.8 ) = 18 0 ∘ + 36.8 7 ∘ ≈ 21 7 ∘ .
Thus,
a ⃗ R = 8 ∠ 217 ∘ \vec{a}_R = 8 \angle 217^\circ a R = 8∠21 7 ∘ .
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Solution Let Q be the origin
( 0 , 0 ) (0,0) ( 0 , 0 ) . Then
P = ( 0 , 12 ) P = (0, 12) P = ( 0 , 12 ) and
R = ( 16 , 0 ) R = (16, 0) R = ( 16 , 0 ) . Given
a ⃗ Q = 0 \vec{a}_Q = 0 a Q = 0 and
a ⃗ R / P = 10 ∠ 180 ∘ = − 10 i ^ \vec{a}_{R/P} = 10 \angle 180^\circ = -10\hat{i} a R / P = 10∠18 0 ∘ = − 10 i ^ .
For a rigid body,
a ⃗ R = a ⃗ Q + α ⃗ × r ⃗ R / Q − ω 2 r ⃗ R / Q \vec{a}_R = \vec{a}_Q + \vec{\alpha} \times \vec{r}_{R/Q} - \omega^2 \vec{r}_{R/Q} a R = a Q + α × r R / Q − ω 2 r R / Q and
a ⃗ P = a ⃗ Q + α ⃗ × r ⃗ P / Q − ω 2 r ⃗ P / Q \vec{a}_P = \vec{a}_Q + \vec{\alpha} \times \vec{r}_{P/Q} - \omega^2 \vec{r}_{P/Q} a P = a Q + α × r P / Q − ω 2 r P / Q .
Subtracting gives
a ⃗ R / P = α ⃗ × r ⃗ R / P − ω 2 r ⃗ R / P \vec{a}_{R/P} = \vec{\alpha} \times \vec{r}_{R/P} - \omega^2 \vec{r}_{R/P} a R / P = α × r R / P − ω 2 r R / P .
With
r ⃗ R / P = 16 i ^ − 12 j ^ \vec{r}_{R/P} = 16\hat{i} - 12\hat{j} r R / P = 16 i ^ − 12 j ^ and
α ⃗ = α k ^ \vec{\alpha} = \alpha\hat{k} α = α k ^ :
− 10 i ^ = α k ^ × ( 16 i ^ − 12 j ^ ) − ω 2 ( 16 i ^ − 12 j ^ ) = ( 12 α − 16 ω 2 ) i ^ + ( 16 α + 12 ω 2 ) j ^ -10\hat{i} = \alpha\hat{k} \times (16\hat{i} - 12\hat{j}) - \omega^2 (16\hat{i} - 12\hat{j}) = (12\alpha - 16\omega^2)\hat{i} + (16\alpha + 12\omega^2)\hat{j} − 10 i ^ = α k ^ × ( 16 i ^ − 12 j ^ ) − ω 2 ( 16 i ^ − 12 j ^ ) = ( 12 α − 16 ω 2 ) i ^ + ( 16 α + 12 ω 2 ) j ^ .
Equating components:
1)
16 α + 12 ω 2 = 0 ⇒ ω 2 = − 4 3 α 16\alpha + 12\omega^2 = 0 \Rightarrow \omega^2 = -\frac{4}{3}\alpha 16 α + 12 ω 2 = 0 ⇒ ω 2 = − 3 4 α 2)
12 α − 16 ω 2 = − 10 ⇒ 12 α − 16 ( − 4 3 α ) = − 10 ⇒ 100 3 α = − 10 ⇒ α = − 0.3 rad/s 2 12\alpha - 16\omega^2 = -10 \Rightarrow 12\alpha - 16(-\frac{4}{3}\alpha) = -10 \Rightarrow \frac{100}{3}\alpha = -10 \Rightarrow \alpha = -0.3 \text{ rad/s}^2 12 α − 16 ω 2 = − 10 ⇒ 12 α − 16 ( − 3 4 α ) = − 10 ⇒ 3 100 α = − 10 ⇒ α = − 0.3 rad/s 2 .
Then
ω 2 = − 4 3 ( − 0.3 ) = 0.4 rad/s 2 \omega^2 = -\frac{4}{3}(-0.3) = 0.4 \text{ rad/s}^2 ω 2 = − 3 4 ( − 0.3 ) = 0.4 rad/s 2 .
Now,
a ⃗ R = α ⃗ × r ⃗ R / Q − ω 2 r ⃗ R / Q = ( − 0.3 k ^ ) × ( 16 i ^ ) − ( 0.4 ) ( 16 i ^ ) = − 4.8 j ^ − 6.4 i ^ \vec{a}_R = \vec{\alpha} \times \vec{r}_{R/Q} - \omega^2 \vec{r}_{R/Q} = (-0.3\hat{k}) \times (16\hat{i}) - (0.4)(16\hat{i}) = -4.8\hat{j} - 6.4\hat{i} a R = α × r R / Q − ω 2 r R / Q = ( − 0.3 k ^ ) × ( 16 i ^ ) − ( 0.4 ) ( 16 i ^ ) = − 4.8 j ^ − 6.4 i ^ .
Magnitude
∣ a ⃗ R ∣ = ( − 6.4 ) 2 + ( − 4.8 ) 2 = 8 m/s 2 |\vec{a}_R| = \sqrt{(-6.4)^2 + (-4.8)^2} = 8 \text{ m/s}^2 ∣ a R ∣ = ( − 6.4 ) 2 + ( − 4.8 ) 2 = 8 m/s 2 .
Angle
θ = 180 ∘ + tan − 1 ( 4.8 6.4 ) = 180 ∘ + 36.87 ∘ ≈ 217 ∘ \theta = 180^\circ + \tan^{-1}(\frac{4.8}{6.4}) = 180^\circ + 36.87^\circ \approx 217^\circ θ = 18 0 ∘ + tan − 1 ( 6.4 4.8 ) = 18 0 ∘ + 36.8 7 ∘ ≈ 21 7 ∘ .
Thus,
a ⃗ R = 8 ∠ 217 ∘ \vec{a}_R = 8 \angle 217^\circ a R = 8∠21 7 ∘ .
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