Physics, asked by sidzcool915, 6 months ago

A bowling ball has a mass of 8.0 kg, a moment of inertia of 3.2 × 10−2 kg·m2, and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 3 m/s, what is its angular speed?

Answers

Answered by Anonymous
4

Explanation:

Total Kinetic energy = translation kinetic energy plus rotationalkinetic energy

TKE = .5mv^2 + .5Iω^2

v = r ω

4 = .1 ω

ω = 40 radians/sec

TKE = .5(7)4^2 + .5 (2.8*10^-2) 40^2

TKE = 56 + 22.4

TKE = 78.4 Joules

Answered by HrishikeshSangha
0

The angular speed of the bowling ball is 30 rad/s.

To find,

Its angular speed.

Given,

A bowling ball has a mass of 8.0 kg,

A moment of inertia of 3.2 × 10−2 kg·m2,

The radius of 0.10 m and

The linear speed of 3 m/s.

Solution,

When an object rolls without slipping, it means that the object's rotational motion and translational motion are coordinated such that the point on the object that is touching the ground (in this case, the bottom of the bowling ball) has zero velocity with respect to the ground. This requires that the angular velocity of the object (the rate at which it rotates) be related to its linear velocity (the rate at which it moves forward) by the equation:

v = ωr

where v is the linear velocity, ω is the angular velocity, and r is the radius of the object.

In this problem, we are given the linear velocity of the bowling ball (3 m/s) and its radius (0.10 m). So we can use the above equation to find the angular velocity:

ω = v / r = 3 m/s / 0.10 m = 30 rad/s

Therefore, the angular velocity of the bowling ball is 30 radians per second.

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