Question

Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one third of ball 2. The kinetic energy of ball 2 is 27.0J. What is the kinetic energy of ball 1?

Answer 1

Answer:

Ball 1 has twice the diameter, half the mass, and one third of ball 2. The kinetic energy of ball 2 is 27.0J. What is the kinetic energy of ball 1

Explanation:

Ball 1 has twice the diameter, half the mass, and one third of ball 2. The kinetic energy of ball 2 is 27.0J. What is the kinetic energy of ball 1

2.0mm falling through a tank of oil at 20C is 6.5m/s. calculate the coefficient of viscosity of the oil at 20C

Answer 2

The kinetic energy of ball 1 is 1.5J.

Given,

Two uniform solid balls- Ball 1 and ball 2, roll at a consistent speed without slipping.

Ball 1 has twice the diameter of ball 2.

Ball 1 has half the mass of ball 2.

Ball 1 has one-third of the speed of ball 2.

The kinetic energy of ball 2 is 27.0J.

To Find,

The kinetic energy of ball 1.

Solution,

We can find the solution to the numerical problem using the following method.

The kinetic energy of a rolling object can be formulated as follows,

K = $\frac{1}{2}I \omega^{2}$+ $\frac{1}{2}$m$v^2$

where I is its rotational inertia, angular speed, mass, and velocity of centre of mass, and m is its mass.

Now assume, the mass, rotational inertia, speed, radius and the diameter of ball 1 is indicated by $m_1 , I_1, v_1,r_1,d_1$. Let m$m_2, I_2, v_2,r_2,d_2$ represent the same for solid ball 2.

As given,

$d_2 =2d_1$.

⇒$2r_2=4r_1$

⇒$r_2=2r_1$.

As they are spheres so, I =$\frac{2}{5}mr^{2}$.

The kinetic energy of ball 1 is,

$K_1=\frac{1}{2}I_1 \omega_1^{2} + \frac{1}{2} m_1v_1^2$

⇒$K_1=\frac{1}{2}(\frac{2}{5}m_1r_1^2) \frac{v_1^2}{r_1^2} } + \frac{1}{2} m_1v_1^2$.

⇒$K_1=\frac{7}{10} m_1v_1^2$.

Similarly for ball 2,

$K_1=\frac{1}{2}I_ 2\omega_2^{2} + \frac{1}{2} m_2v_2^2$

⇒$K_1=\frac{1}{2}(\frac{2}{5}m_2r_2^2) \frac{v_2^2}{r_2^2} } + \frac{1}{2} m_2v_2^2$.

⇒$K_1=\frac{7}{10} m_2v_2^2$.

We have $m_1=\frac{m_2}{2} and v_1=\frac{v_2}{3}$.

By putting those values in $K_1$ we get,

⇒$K_1=\frac{7}{10} \frac{m_2}{2} (\frac{v_2}{3} )^2$.

⇒$K_1=\frac{1}{18}( \frac{7}{10} m_2v_2^2)$.

⇒$K_1=\frac{1}{18}K_2$.

(As the kinetic energy of ball 2,$K_2$ is 27.0J.)

⇒$K_1=\frac{1}{18}.27$=1.5 .

Hence, the kinetic energy of ball 1 is 1.5J.

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