An 8.00-cm-diameter, 360 g solid sphere is released from rest

Concepcion Hale

Concepcion Hale

Answered question

2022-01-16

An 8.00-cm-diameter, 360 g solid sphere is released from rest at the top of a 1.8-m-long, 17.0 incline. It rolls, without slipping, to the bottom. What is the sphere’s angular velocity at the bottom of the incline? What fraction of its kinetic energy is rotational?

Answer & Explanation

poleglit3

poleglit3

Beginner2022-01-17Added 32 answers

Step 1
diameter(d)=8cm
=0.08m
mass(m)=360g
=0.36kg
Len>h(L)=1.8m
Angle of incline (θ)=17
Step 2
a) Using conservation of energy:
potential energy=kinetic energy
m×g×h=(12×m×v2)+(12×I×ω2)
using, v=r×ω
m×g×h=(12×m×(r×ω)2)+(12×(25×m×r2)×ω2)
9.8×1.8×sin17=0.7×(0.04)2×ω24604.85=ω2
67.8rads=ω
(Angular velocity at bottom)
b) rotational kinetic energytotal kinetic energy=(12×I×ω2)(12×m×v2)+(12×I×ω2)
=(12×m×r2×ω2)0.7×m×r2×ω2
=0.714
Jeffery Autrey

Jeffery Autrey

Beginner2022-01-18Added 35 answers

Answer:
a) 88.1 rad/s
b) 0.286
Explanation:
given information:
diameter, d=8cm=0.08m
sphere's mass, m=400g=0.4kg
the distance from rest to the tip, h=2.1m
incline angle, θ=25
a) What is the sphere's angular velocity at the bottom of the incline?
mg(hsinθ)=12Iω2+12mv2
I of solid sphere =25mr2, thus
mg(hsinθ)=12(25mr2)ω2+12mv2, now we can remove the mass
ghsinθ=15r2ω2+12v2
ω=vr,v=ωr
so,
ghsinθ=15r2ω2+12(ωr)2
ghsinθ=(710)r2ω2
ω2=10ghsinθ7r2
ω=10ghsinθ7r2
=10(9.8)(2.1)sin257(0.04)2
=88.1rads
b) What fraction of its kinetic energy(KE) is rotational?
fraction of its kinetic energy =rotational KE/total KE
total KE=total potential energy
=mghsinθ
=0.4×9.8×2.1sin25
=3.48J
rotational KE=12Iω2
=15mr2ω2
=150.4(0.04)2(88.1)2
=0.99
fraction of its KE

alenahelenash

alenahelenash

Expert2022-01-23Added 556 answers

An 8.6cm diameter , 320 g solid sphere is released from rest at the top of a 1.6 m long, 19 incline with no slipping. What is the sphere’s angular velocity at the bottom of the incline? What fraction of the energy is rotational? The total energy of a body undergoing rolling motion is, E=ET+ER, where ET and ER are the translational kinetic energy and the rotational kinetic energy respectively. ET=12mv2, where m and v are the mass and velocity respectively. ER=12Iω2, where I and ω are the moment of inertia and the angular velocity respectively. I=mK2 and ω=vR, where K and R are the radius of gyration and the radius of the sphere respectively. ER=12Iω2=12mK2(vR)2. E=12mv2+12mK2(vR)2=12mv2(1+K2R2). This kinetic energy has been obtained from the potential energy that the sphere was having just before it started rolling. This potential energy is E=mgh=mgssinθ, where g,h,s and θ are acceleration due to gravity, vertical height, slant height and the angle of inclination. 12mv2(1+K2R2)=mgssinθ. v=2gssinθ1+K2R2 For a solid sphere, K=25RK2R2=25 v=2gssinθ1+K2R2=10gssinθ7. In this particular case, it is given that s=1.6m,R=4.3×102m,m=0.32kg and θ=19 v=10×9.8×1.6×sin19710×9.8×1.6×0.325672.7m/s Angular velocity,ω=vR=2.74.3×102=62.79radian/s. The fraction of the energy that is rotation is ERE. =12mv2(K2R2)12mv2(1+K2R2)=K2R21+K2R2=251+25=27.

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