Question

1) A solid sphere of radius = 10 cm and mass = 2 kg is going down an inclined plane of height = 5 m. The angle of the inclined plane = 35 degrees.

a) What is the final Total kinetic energy of the sphere? 98J

b) Calculate the final Vcm and angular of the sphere and the Total velocity at the top point. Vcm=8.37m/s w=83.67 rads/s Vt=16.73m/s

c) What are the final Linear (translational) and Rotational kinetic energies of the sphere? KE(t)=70 J KE(rot)=28J

d) Calculate the acceleration of the center of mass of the sphere and the Total acceleration at the top point and the radial acceleration (alpha) of the sphere. a(cm)=4.015m/s^2, a(t)=8.03m/s^2, angular acceleration (alpha)=40.15rad/s^2

e) How many seconds will it take the sphere to reach the bottom? t=2.084 seconds

f) What is the magnitude of fs. What is the minimum coefficient of static friction required for pure rolling motion? fs=3.21N coefficient of static friction=0.20

I need help with the step-by-step.

Answer 1

Given ,

mass of the sphere ,m= 2kg

height of the inclined plane,h = 5m

Radius of the sphere,r= 10cm

g=9.8m/s²

a) The total kinetci energy experienced by the sphere during roling down from an incline is givenby

potential energy at rest = Translational Kinetic energy(KE_T)+ Rotational Kinetic energy (KE_R)

ie,

$mgh=\frac{1}{2}mv^{2} +\frac{1}{2}Iw^{2}$

where v= velocity of the sphere while roling

w= angular velocity

I= moment of inertia

For a solid sphere the moment of inertia is given as , $I=\frac{2}{5}m r^{2}$

Therefore,

$2gh= v^{2}+\frac{2}{5} r^{2}(\frac{v}{r})^{2}$

$2gh= v^{2}+\frac{2}{5} v^{2}$

$2gh= \frac{7}{5}v^{2}$

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

$v= \sqrt{\frac{10}{7}\times 9.8 \times 5}$

=8.37 m/s

Angular velocity

, $w=\frac{v}{r}$

$w=\frac{8.367}{0.1}$

=83.67 rad/s

The total kinectic enery= K_T + K_R

The Translational Kinectic energy,

$K_{T}=\frac{1}{2} m v^{2}$

$K_{T}=\frac{1}{2} \times 2\times 8.37^{2}$

= 70 J

The rotational Kinectic energy,

$K_{R}=\frac{1}{2} I w^{2}$

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

$K_{R} = \frac{1}{2}(\frac{2}{5} \times 2\times 0.1^{2} \times( \frac{8.37}{0.1})^{2})$

= 28 J

The to tal Kinectic energy= 28+70= 98 J

In question $\mathbf{v= v_{cm}=8.37 m/s}$

Total velocity at top = 2 v = 2 X 8.37 = 16.74 m/s

a) Total Kinectic Energy = 98J

b) Angular velocity, w= 83.67 rad/s

   Total velocity V_t = 16.74 m/s

final V_cm = 8.37 m/s

c)Rotational Kinectic energy ,K_R = 28 J

Translational Kinectic energy, K_T =70 J