A 42cm diameter wheel consists of a rim made from a rigid plastic, with linear d

Question

A 42cm diameter wheel consists of a rim made from a rigid plastic, with linear density of 25.0g/cm and 6 spokes (hint: 6 slender rods of linear density), released at the top of a 58.0m. high hill. How fast is the wheel rolling (i.e. angular velocity) at the bottom of the hill? A 42cm diameter wheel consists of a rim made from a rigid plastic, with linear density of 25.0g/cm and 6 spokes (hint: 6 slender rods of linear density), released at the top of a 58.0m. high hill. How fast is the wheel rolling (i.e. angular velocity) at the bottom of the hill? A 42cm diameter wheel consists of a rim made from a rigid plastic, with linear density of 25.0g/cm and 6 spokes (hint: 6 slender rods of linear density), released at the top of a 58.0m. high hill. How fast is the wheel rolling (i.e. angular velocity) at the bottom of the hill?

Explanation / Answer

diameter of the wheel, d=42cm

radius of the wheel, r=0.21m

no of spokes is 6

linear density of spokes is u=25g/cm

height of the hill, h=58m

by using conservation of enegry,

m*g*h=1/2*m*v^2+1/2*I*w^2

here,

I=I_rim + I_spokes

I=m_rim*r^2+6*(1/3*m_spokes*r^2)

and

m_rim=u*(2pi*r)

m_spoke=u*r

now,

I=u*2pi*r^3+2*u*r^3

I=2*u*r^3(pi+1)

and

m*g*h=1/2*m*v^2+1/2*I*w^2

m*g*h=1/2*m*(r*w)^2+1/2*(2*u*r^3(pi+1)*w^2

m*g*h=1/2*m*r^2*w^2 + u*r^3(pi+1)*w^2

here,

m=u*(2r*(pi+3))

then,

m*g*h=1/2*m*r^2*w^2 + u*r^3(pi+1)*w^2

u*(2r*(pi+3))*g*h=1/2*u*(2r*(pi+3))^2*r^2*w^2 +u*r^3(pi+1)*w^2

2r*(pi+3)*g*h=r*(pi+3)^2*r^2*w^2 + r^3(pi+1)*w^2

====>

w=sqrt((pi+3)*g*h)/(r^2*(pi+2))

w=sqrt((pi+3)*9.8*58)/(0.21^2*(pi+2))

w=260.57 rad/sec ----->

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