Question

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at 33.0°. It momentarily stops when it has rolled 1.70 m along the ramp. What was its initial speed?

Answer 1

Answer:

v = 3.6 m/s

Explanation:

given,

Inclination of the ramp, θ = 33⁰

Distance of the rope, d = 1.70 m

initial speed = ?

using conservation of energy

$\dfrac{1}{2}mv^2 + \dfrac{1}{2}I_{com}\omega^2 = m g h$

h = d sinθ  

h = 1.7 x sin 33° = 0.926 m

moment of inertia of the solid ball

$I_{com}= \dfrac{2}{5}MR^2$  

we know, v = r ω

$\dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{2}{5}mR^2\times \dfrac{v^2}{R^2}= m g h$

$\dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2 = mgh$

$gh= \dfrac{7}{10}v^2$

$v = \sqrt{\dfrac{10}{7}gh}$

$v = \sqrt{\dfrac{10}{7}\times 9.8\times 0.926}$

v = 3.6 m/s

Hence, initial speed of the solid ball = 3.6 m/s