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Len [333]
4 years ago
13

A 50 gram yo-yo is released and travels vertically downward, rolling without slipping. Knowing that the yo-yo’s moment of inerti

a is 9 x 10-5 kg m2 and the radius of its center spindle is 0.7 cm,
find
A) the torque involved and
B) the angular acceleration of the yo-yo.
Physics
1 answer:
morpeh [17]4 years ago
6 0

Answer:

Explanation:

mass, m = 50 g

moment of inertia, I = 9 x 106-5 kg m^2

radius, r = 0.7 cm

(a) As it moving downwards

Let T be the tension in the string

T = m (g + a)   .... (1)

where, a be the acceleration

τ = I α = T r

α =  a / r

So, I x a / r = T x r

a = T r^2 / I

Substitute in equation (1) we get

a = m (g + a) r^2 / I

a = mgr^2 / (I - mr^2)

a = 0.050 x 9.8 x 0.007 x 0.007 / (9 x 10^-5 - 0.050 x 0.007 x 0.007)

a = 2.401 x 10^-5 / (87.55 x 10^-6)

a = 0.274 m/s^2

τ = I x α = I x a / r

τ = 9 x 10^-5 x 0.274 / 0.007

τ = 3.52 x 10^-3 Nm

(b) α = a / r

α = 0.274 / 0.007 = 39.14 rad/s^2

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Answer:

≅50°

Explanation:

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Now luckily we are given everything we need to solve (or you found the info before posting here):

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At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.

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i hope this helps.

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