Question

A $200g$ block on a $50-cm$-long string swings in a circle on a horizontal, frictionless table at $75rpm$.

a. What is the speed of the block?

b. What is the tension in the string?

Short answer

a). The speed of the block is $3.9m/s$.

b). The tension in the string is$6.2N$.

Step-by-step solution

Step 1:Given Information (Part a)

A$200g$ block on a $50-cm$-long string swings in a circle on a horizontal, frictionless table at $75rpm$.

Explanation (Part a)

The block velocity is tangent to the circle of motion, and its acceleration is called the centripetal acceleration and it points toward the center of the circle. The block has angular velocity $\omega$and a tangential speed $v_T$and they are related to each other by

$v_T=r\omega$

Where $r$is the radius of the circle. The angular velocity is given by $\omega =75\mathrm{rpm}=75\mathrm{rev}/\min$. In$\mathrm{SI}$ units, the angular frequency has unit radians per second ( $\mathrm{rad}/\mathrm{s})$. So, let us convert the unit from and $1\min =60\mathrm{second}\mathrm{s}$

$\omega =\left(75\frac{\mathrm{rev}}{\min }\right)\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)=7.85\mathrm{rad}/\mathrm{s}$

Now, plug the values for $\omega$and $r$into equation (1) to get $v_T$

$v_T=\omega r=(7.85\mathrm{rad}/\mathrm{s})(0.50\mathrm{m})=3.925\mathrm{m}/\mathrm{s}\approx 3.9\mathrm{m}/\mathrm{s}$

Final Answer (Part a)

The speed of the block is$3.9m/s$.

Given Information (Part b)

A $200g$ block on a $50-cm$-long string swings in a circle on a horizontal, frictionless table at $75rpm$.

Explanation (Part b)

This acceleration of the block is given by equation (8.4) in the form

$\overrightarrow{a}=\frac{v^2}{r}$

Where $r$ is the radius of the circle and $v$ is the velocity of the block that we are calculated in part (a). At a uniform circular motion, the velocity vector has a tangential component and the acceleration vector has a radial component. From Newton 's first law, the block has a net force exerted on it as it doesn't move with a constant velocity. So, using the expression of the acceleration from equation (2), we get the net force by

$\overrightarrow{F_{\text{net}}}=m\overrightarrow{a}=\frac{m{v}^2}{r}$

Explanation (Part b)

This force has a direction toward the center of the circular motion. The block moves in a circle with a radius $r=0.50m$. Now, we plug the values for $m,v$and $r$into equation (3) to get $F_{\text{net}}$

$F_{\text{net}}=\frac{m{v}^2}{r}$

$=\frac{(0.200\mathrm{kg})(3.925\mathrm{m}/\mathrm{s}{)}^2}{0.50\mathrm{m}}$

$=6.16\mathrm{N}$

$\approx 6.2\mathrm{N}$

This force is an identifiable agent and it is one of our familiar forces such as friction, normal or tension forces. The block moves on the table in a circular motion, there is no friction force because the table is frictionless, also there is no horizontally normal force, so the tension force in the wire causes this force. Therefore, the tension force will be

$T=6.2\mathrm{N}$

Final Answer (Part b)

The tension force will be$6.2N$.