Answer:
r = 0.0173 m = 1.73 cm
Explanation:
Here, the centripetal force of the block will be providing the required breaking tension in the string:
[tex]Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\[/tex]
where,
r = radius = ?
m = mass of block = 0.13 kg
v = tangential spee of block = 4 m/s
T = Breaking Strength = 30 N
Therefore,
[tex]r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}[/tex]
r = 0.0173 m = 1.73 cm
When finding the radius of the string at the point it breaks, the tangential
velocity is assumed to be constant.
Reasons:
The mass of the small block, m = 0.130 kg
Initial radius of the circle of rotation = 0.800 m
Tangential velocity, v = 4.00 m/s
The radius of the path of rotation is reduced as the string is pulled
Breaking strength of the string = 30.0 N
Required:
The radius of the circle when the string brakes
Solution:
[tex]Centripetal \ force = \dfrac{m \cdot v^2}{r}[/tex]
Where;
r = The radius of the circle of rotation
When the string brakes, w have;
Centripetal force = Breaking strength of the string = 30.0 N
Which gives;
[tex]\displaystyle r = \mathbf{\dfrac{m \cdot v^2}{Centrifugal \ force}} = \frac{0.130 \times 4^2}{30} =6.9\overline 3 \times 10^{-2}[/tex]
The radius of the circle when, the string breaks r = [tex]\underline{6.9\overline 3 \times 10^{-2}} \ m[/tex]
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