Question

In: Physics

A toy top with a spool of diameter 5.0cm has a moment of inertia of 3.0x10^-5kg x m^2 about its rotation axis.

A toy top with a spool of diameter 5.0cm has a moment of inertia of 3.0x10^-5kg x m^2 about its rotation axis. To get the top spinning, its string is pulled with a tension of .30 N. How long does it take for the top to complete the first five revolutions? The string is long enough that it is wrapped around the top more than five turns

Solutions

Expert Solution

Given that

The diameter of the \(s p o o l, d=5.0 \mathrm{~cm}\)

So the radius of the spool, \(r=2.5 \mathrm{~cm}\)

$$ =2.5 \times 10^{-2} \mathrm{~m} $$

Moment of Inertia, \(I=3 \times 10^{-5} \mathrm{kgm}^{2}\) Tension force, \(T=0.30 \mathrm{~N}\)

We have a formula for torque as

$$ \tau=l \alpha $$

and also we have, \(\tau=r F\)

From the above two equations,

$$ \begin{array}{l} l \alpha=r F \\ \alpha=\frac{r F}{I} \\ \alpha=\frac{\left(2.5 \times 10^{-2} \mathrm{~m}\right)(0.30 \mathrm{~N})}{\left(3 \times 10^{-5} \mathrm{kgm}^{2}\right)} \\ \alpha=250 \mathrm{rad} / \mathrm{s}^{2} \end{array} $$

Since the spool is starting from rest its initial vecoity is zero.

So we have formula as

$$ \theta=\frac{1}{2} \alpha t^{2} $$

Here the angular di splacement, \(\theta=(5 \times 2 \pi)\) rad

From the above the expression for time is given by

$$ \begin{aligned} t &=\sqrt{\frac{2 \theta}{\alpha}} \\ &=\sqrt{\frac{2(10 \pi)}{250}} \\ &=0.501 \end{aligned} $$

So the required time is given by \(\overline{t=0.5 \mathrm{~s}}\)


t = 0.5s.

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