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Physics Question on torque

The position vector r⃗\vec{r} of a particle of mass mm is given by the following equation r⃗(t)=αt3i^+βt2j^\vec{r}(t)=\alpha t^{3} \hat{i}+\beta t^{2} \hat{j} where α=103 ms−3,β=5 ms−2\alpha=\frac{10}{3} \,ms ^{-3}, \beta=5\, ms ^{-2} and m=0.1 kgm =0.1 \,kg. At t=1 st =1\, s, which of the following statement (s) is (are) true about the particle?

A

The velocity v→\overrightarrow{ v } is given by v→=(10i^+10j^)ms−1\overrightarrow{ v }=(10 \hat{ i }+10 \hat{ j }) ms ^{-1}

B

The angular momentum L→\overrightarrow{ L } with respect to the origin is given by L→=−(53)k^Nms\overrightarrow{ L }=-\left(\frac{5}{3}\right) \hat{ k } N ms

C

The force F→\overrightarrow{ F } is given by F→=(i^+2j^)N\overrightarrow{ F }=(\hat{ i }+2 \hat{ j }) N

D

The torque τ⃗\vec{\tau} with respect to the origin is given by τ⃗=−(203)k^Nm\vec{\tau}=-\left(\frac{20}{3}\right) \hat{k} Nm

Answer

The torque τ⃗\vec{\tau} with respect to the origin is given by τ⃗=−(203)k^Nm\vec{\tau}=-\left(\frac{20}{3}\right) \hat{k} Nm

Explanation

Solution

r⃗(t)=αt3i^+βt2j^\vec{r}(t)=\alpha t^{3} \hat{i}+\beta t^{2} \hat{j}
α=103ms−3,β=5 ms−2\alpha=\frac{10}{3} ms ^{-3}, \beta=5\, ms ^{-2} and m=0.1 kg.t=1 sm =0.1\, kg . t =1\, s, v→=3αt2i^+2βtj^\overrightarrow{ v }=3 \alpha t ^{2} \hat{ i }+2 \beta t \hat{ j }
(A) (v→)r=1=3αi^+2βj^(\overrightarrow{ v })_{ r =1}=3 \alpha \hat{ i }+2 \beta \hat{ j }
=3×103i^+2×5j^=3 \times \frac{10}{3} \hat{ i }+2 \times 5 \hat{ j }
=10i^+10j^=10 \hat{ i }+10 \hat{ j }
(B) L⃗=r⃗×P→=m(r→×v→)\vec{L} =\vec{r} \times \overrightarrow{ P }= m (\overrightarrow{ r } \times \overrightarrow{ v })
=0.1[(αt3i^+βt2j^)×(10i^+10j^)]=0.1\left[\left(\alpha t ^{3} \hat{i}+\beta t ^{2} \hat{ j }\right) \times(10 \hat{ i }+10 \hat{ j })\right]
=0.1[(10αt3(k^)−10βt2(k^)]=0.1\left[\left(10 \alpha t ^{3}(\hat{ k })-10 \beta t ^{2}(\hat{ k })\right]\right.
=0.1[10×103×1k^−10×5(1)2k^]=0.1\left[10 \times \frac{10}{3} \times 1 \hat{ k }-10 \times 5(1)^{2} \hat{ k }\right]
=0.1[1003−50]k^=0.1\left[\frac{100}{3}-50\right] \hat{ k }
=−53(k^)=-\frac{5}{3}(\hat{k})
(C) F→=ma→=m[6αti^+2βj^]\overrightarrow{ F } = m \overrightarrow{ a }= m [6 \alpha t \hat{ i }+2 \beta \hat{ j }]
0.1[6×103×1i^+2×5j^]0.1\left[6 \times \frac{10}{3} \times 1 \hat{ i }+2 \times 5 \hat{ j }\right]
=[2i^+j^]=[2 \hat{ i }+\hat{ j }]
(D) τ⃗=r→×F→\vec{\tau} =\overrightarrow{ r } \times \overrightarrow{ F }
=(αt3i^+βt2j^)×[2i^+j^]=\left(\alpha t ^{3} \hat{ i }+\beta t ^{2} \hat{ j }\right) \times[2 \hat{ i }+\hat{ j }]
=(103i^+5j^)×[2i^+j^]=\left(\frac{10}{3} \hat{ i }+5 \hat{ j }\right) \times[2 \hat{ i }+\hat{ j }]
=103k^−10k^=\frac{10}{3} \hat{ k }-10 \hat{ k }
=−203k^=\frac{-20}{3} \hat{ k }