Q5-43E - University Physics with Modern Physics

Question

A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds60.0 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. 

(a) Draw a free body diagram of the stone. 

(b) Find the maximum speed the stone can attain without the string breaking.

Step-by-Step Solution

Answer

(a)  

(b) The maximum speed the stone can attain is,$8.21\text{\hspace{0.17em}m/s}$ .

Step 1: Identification of given data

The given data can be listed below as,

• The mass of stone is, $m=0.80\text{\hspace{0.17em}kg}$.
• The length of the string is,$r=0.90\text{\hspace{0.17em}m}$ .
• The tensile strength of the string is,$T=60.0\text{\hspace{0.17em}N}$ .

Step 2: Significance of tension

Tension is a force which is carried by a flexible medium such as rope, string, and cable, etc. Tension force acts along the length of the rope, string, cable, etc.

Step 3: Determination of free body diagrams of the stone

Part (a)

 

The free body diagrams of the stone can be expressed as,

Step 4: Determination of maximum speed the stone can attain

Part (b)

 

The relation of tension with centripetal force can be expressed as,

 $T=\frac{m{v}^2}{r}$

Here $m$ is the mass of the stone, $r$ is the length of the string, $v$ is the maximum speed of the stone.

 

Substitute 0.80 kg for $m$ , 0.90 m for $r$ , 60.0 N for $T$ in the above equation,

 

$\begin{array}{c}60.0\text{\hspace{0.17em}N}=\frac{0.80\text{\hspace{0.17em}kg}\times v^2}{0.90\text{\hspace{0.17em}m}}\\ v=\sqrt{\frac{0.90\text{\hspace{0.17em}m}\times 60.0\text{\hspace{0.17em}N}}{0.80\text{\hspace{0.17em}kg}}}\\ =8.21\text{\hspace{0.17em}m/s}\end{array}$

Hence, the required maximum speed is,$8.21\text{\hspace{0.17em}m/s}$ .