A 2.5-kg solid sphere starts from rest at the top of a ramp and rolls without slipping. Calculate its speed wh

A 2.5-kg solid sphere starts from rest at the top of a ramp and rolls without slipping. Calculate its speed when it reaches the bottom of the ramp. The ramp is 0.79 m high and 5.0 m long.

As there is no slipping,
The kinetic energy due to rotation is two fifth of the linear kinetic energy.

M g h = 0.5 M U² + 0.5 (0.4 M R²) U² / R²

g h = 0.5* U² + 0.5*0.4 U²
9.8* 0.79 = 0.5* U² + 0.5*0.4 U²
U = 3.33 m/s
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If you want to see the derivation of the Pearlsawme Answer, here it is:


A sphere "falling" a given height h (by rolling down the ramp) will convert all its potential energy PE to kinetic KE by the time it reaches the bottom:

(1) KE = PE = m * g * h

The kinetic energy at the bottom KEf is made up of a translational KEt plus a rotational KEr:

(2) KEf = KEt + KEr

where
(3) KEt = ½ m * v²
and
(4) KEr = ½ I * ω²

where
v is the final translational velocity and ω is the final angular (rotational) speed
and
(5) ω = v / r . . . where r is the radius of the sphere.

I is the moment of inertia of a solid sphere:

(6) I = 0.4 * m * r² . . . . [see Source 1]

Substituting the above into (2) to get an expression for KEf:

(7) KEf = KEt + KEr

= (½ m * v²) + (½ I * ω²)

= (½ m * (r * ω)²) + (½ (0.4 * m * r²) * ω²). . . [substitutions from (5) and (6)]

= m * ½ (r² * ω² + 0.4 * r² * ω²)

= 0.7 * m * r² * ω²

= 0.7 * m * v² . . . [substitution from (5) again]

Therefore, recalling (1) and solving for the final velocity v:

(8) v = √ (PE / (0.7 * m) )

= √ (m * g * h / (0.7 * m) )

= √ (g * h / 0.7 )

= sqrt (9.8 * 0.79 / 0.7 )

= 3.3m/s


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