A disc without slipping rolls down a hill height 10.0 m. If the disc starts from rest at the top of hill, what is its speed at the bottom?
A disc without slipping rolls down a hill of height 10m. If the disc starts form rest at the top of the hill, its speed at the bottom is:
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To find the speed of the disc at the bottom of the hill, we can use the principle of conservation of mechanical energy.
1. First, let's calculate the potential energy (PE) of the disc at the top of the hill. The potential energy is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the height is 10.0 m.
2. Next, we need to find the kinetic energy (KE) of the disc at the bottom of the hill. Since the disc is rolling without slipping, its total kinetic energy is the sum of the translational kinetic energy (KE_t) and the rotational kinetic energy (KE_r). The translational kinetic energy is given by the formula KE_t = 0.5mv^2, where m is the mass and v is the linear velocity. The rotational kinetic energy is given by the formula KE_r = 0.5Iω^2, where I is the moment of inertia and ω is the angular velocity.
3. At the top of the hill, the disc is at rest, so its total mechanical energy is equal to its potential energy. Therefore, the total mechanical energy (E) is given by E = PE = mgh.
4. At the bottom of the hill, the total mechanical energy is the sum of the translational and rotational kinetic energies. Therefore, E = KE_t + KE_r.
5. Since there is no slipping, the linear velocity (v) of the disc at the bottom of the hill is related to the angular velocity (ω) by the equation v = rω, where r is the radius of the disc.
6. Now, we substitute the expressions for KE_t and KE_r into the equation E = KE_t + KE_r and rearrange to solve for v.
E = 0.5mv^2 + 0.5Iω^2
mgh = 0.5mv^2 + 0.5I(ω^2)
Since v = rω,
mgh = 0.5mv^2 + 0.5I(v^2/r^2)
Simplifying, we can cancel out the mass 'm' from both sides of the equation:
gh = 0.5v^2 + 0.5I(v^2/r^2)
gh = 0.5v^2(1 + I/r^2)
Simplifying further, we can substitute I = 1/2MR^2, where M is the mass of the disc and R is its radius:
gh = 0.5v^2(1 + (1/2MR^2)/r^2)
gh = 0.5v^2(1 + 1/2(Mr^2)/(r^2))
gh = 0.5v^2(1 + 1/2(R^2/r^2))
gh = 0.5v^2(1 + 1/2(R/r)^2)
Therefore, we can solve for v:
v^2 = (2gh) / (1 + 1/2(R/r)^2)
v = sqrt((2gh) / (1 + 1/2(R/r)^2))
Substituting the given values of h (10.0 m) and R/r (assuming the disc is solid) into the equation, we can calculate the final speed of the disc at the bottom of the hill.
= 1/2(I)w^2 + 1/2 m v^2= 1/2 (Mr^2)(v^2/r) + 1/2 M v^2
Sir ma Dard na karo Ziyda Phaar karr
final KE=initial PE
where initial PE=mgh=mass*9.8*10
final KE = rolling KE + translational KE
= 1/2(I)w^2 + 1/2 m v^2= 1/2 (Mr^2)(v^2/r) + 1/2 M v^2
put all in this equation
final KE=initial PE
and solve for veloicy v.