contestada

Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.35 m high. assume it starts from rest and rolls without slipping.

Respuesta :

The total kinetic energy will be equivalent to the potential energy;
The total kinetic energy is given by the sum of transitional kinetic energy and the rotational kinetic energy.
Rotational kinetic energy
I= 1/2mR², but rotational energy =1/2 I w² 
Thus  rotational energy is equal to 1/4 mR²w², which is equivalent to 
 = 1/4 mv² since v= wR
Thus;
Total kinetic energy = 1/2 mv² + 1/4 mv²
hence;
 Mgh = 3/4 mv²
  gh = 3/4 v² ( assuming g = 9.81 m/s²)
   72.1035 = 3/4 v²
v = 9.805
    = 9.805 m/s
shubhamchouhanvrVT

The translational speed of a cylinder when it reaches the foot of an incline 7.35 m high will be   [tex]V=9.80\frac{m}{s}[/tex]

What will be the translation speed of the cylinder?

It is given that

Height = 7.35 m

Now we know that rotational kinetic energy is given as

[tex]K_r=\dfrac{1}{2} Iw^2[/tex]

Here [tex]I=\dfrac{1}{2} mR^2[/tex]

[tex]K_R= \dfrac{1}{2} \dfrac{1}{2} mR^2 w^2[/tex]

[tex]K_R= \dfrac{1}{4} mR^2w^2[/tex]

As we know that  linear  velocity is given as

[tex]V=wR[/tex]

[tex]K_R= \dfrac{1}{4} mV^2[/tex]

So the total kinetic energy will be the sum of the linear kinetic energy and the rotational kinetic energy.

[tex]K_T=K_L+K_R[/tex]

[tex]K_T= \dfrac{1}{2} mV^2+ \dfrac{1}{4} mV^2= \dfrac{3}{4} mV^2[/tex]

Now from conservation of energy

[tex]PE= K_T[/tex]

[tex]mgh= \dfrac{3}{4} mV^2[/tex]

[tex]gh= \dfrac{3}{4} V^2[/tex]

[tex]\dfrac{9.8\times7.35\times 4}{3} =V^2[/tex]

[tex]V=9.80 \dfrac{m}{s}[/tex]

Thus the translational speed of a cylinder when it reaches the foot of an incline 7.35 m high will be   [tex]V=9.80\frac{m}{s}[/tex]

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