Question

A uniform, solid sphere of radius 4.00 cm and mass 3.75 kg starts with a purely translational speed of 3.75 m/s at the top of an inclined plane. The surface of the incline is 2.75 m long, and is tilted at an angle of 30.0∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed v2 at the bottom of the ramp.

Answer 1

Answer:

The velocity or speed V2 at the bottom of the ramp is

$vf=5.129 \frac{m}{s}$

Explanation:

Calculated the velocity of the sphere at the bottom of the ramp using the law of conservation of energy

$E_{p}= E_{k}$

$\frac{1}{2}*m*vf^2+\frac{1}{2}I*w^2=\frac{1}{2}*m*vi^2+m*g*h$

But Inertia can be use in value using in the initial equation so:

$I=\frac{2}{5}*m*r^{2} \\ w=\frac{v}{r}$

$\frac{1}{2}*m*vf^2+\frac{1}{5}*m*vf^2=\frac{1}{2}*m*vi^2+m*g*h$

$\frac{7}{10}*m*vf^2=\frac{1}{2}*m*vi^2+m*g*h$

So replacing also can get factor m from all terms an eliminated 'm'

$vf^{2}=(\frac{5}{7})vi^2+(\frac{10}{7})*g*h \\vf=\sqrt{(\frac{5}{7})vi^2+(\frac{10}{7})*g*h}$

To find 'h'

$h=L*sin(\alpha) \\h=2.75*sin(25)=1.162m\\h=1.162 m$

$vf=\sqrt{\frac{5}{7}*(3.75\frac{m}{s})^{2}+\frac{10}{7}*9.8\frac{m}{s^{2}}*1.1622m}\\ vf=5.129 \frac{m}{s}$

Answer 2

The sphere's final translational speed $V_2$ at the bottom of the ramp is equal to 5.41 m/s.

Given the following data:

• Radius of sphere = 4.00 cm

• Mass of sphere = 3.75 kg

• Speed = 3.75 m/s

• Length = 2.75 m

• Angle of inclination = 30.0°

To calculate the sphere's final translational speed $V_2$ at the bottom of the ramp, we would apply the law of conservation of energy and :

Total kinetic energy at the top + Work done by gravity = Total kinetic energy at the bottom.

Note: Total kinetic energy at the bottom = Final translational K.E + Rotational K.E

$K.E_{top} + W_{g} = K.E_{trans} + R.E\\\\\frac{1}{2} mv^2_1 + mgdsin\theta = \frac{1}{2} mv^2_2 + \frac{1}{2} I \omega^2$   ....equation 1.

For a sphere:

$I = \frac{2}{5} mr^2$   ...equation 2.

For angular speed:

$\omega = \frac{v}{r}$    ....equation 3.

Substituting the value I and $\omega$ into eqn. 1, we have:

$\frac{1}{2} mv^2_1 + mgdsin\theta = \frac{1}{2} mv^2_2 + \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v_2}{r} )^2\\\\\frac{1}{2} mv^2_1 + mgdsin\theta = \frac{1}{2} mv^2_2 + \frac{1}{5} ( mv_2^2)\\\\\frac{1}{2} mv^2_1 + mgdsin\theta = \frac{7}{10} mv^2_2$

Multiplying both sides by $\frac{10}{7m}$, we have:

$\frac{5}{7} V_1^2+\frac{10}{7} gdsin\theta = V_2^2\\\\V_2=\sqrt{\frac{5}{7} V_1^2+\frac{10}{7} gdsin\theta}$

Substituting the given parameters into the formula, we have;

$V_2=\sqrt{\frac{5}{7} \times(3.75)^2+\frac{10}{7}\times (9.8 \times 2.75 sin(30))} \\\\V_2=\sqrt{10.05+19.25} \\\\V_2=\sqrt{29.3} \\\\V_2 = 5.41\;m/s$

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