A particle of mass $10 \mathrm{~g}$ moves along a circle of radius $6.4 \mathrm{~cm}$ with a constant…

1. 0.6 $m/s^2$
2. 0.4 $m/s^2$
3. 0.1 $m/s^2$
4. 0.3 $m/s^2$

Solution

Consider the following diagram: The tangential acceleration is given by, $\begin{aligned} & a_t=\frac{\mathrm{d} v}{\mathrm{~d} t}=\text { Constant } \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=a_t \\ & \Rightarrow \frac{\mathrm{d} s}{\mathrm{~d} t} \cdot \frac{\mathrm{d} v}{\mathrm{~d} s}=a t \\ & \Rightarrow v=\frac{\mathrm{d} v}{\mathrm{~d} t}=a_t \\ & \Rightarrow \int_0^v \frac{v^2}{2} \mathrm{~d} v=\int_0^{(4 \pi r)} a_t \mathrm{~d} s \end{aligned}$ We know, Now, kinetic energy is given by, $K E=\frac{m v^2}{2}$ Now, using eq ${ }^{\mathrm{n}}(1) \&(2)$ $\begin{aligned} & \left(\frac{2 K E}{m}\right)=(8 \pi r) a_t \\ & \Rightarrow a_t=\left(\frac{K E}{(4 \pi r) m}\right) \end{aligned}$ Given, $K E=8 \times 10^{-4} \mathrm{~J}, r=6.4 \times 10^{-2} \mathrm{~m}$ and $m=1 \times 10^{-2} \mathrm{~kg}$. $\Rightarrow a_t=\frac{\left(8 \times 10^{-4}\right) \mathrm{J}}{4 \pi\left(6.4 \times 10^{-2} \mathrm{~m}\right)\left(10^{-2} \mathrm{~kg}\right)}=0.1 \mathrm{~m} / \mathrm{s}^2$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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