The same force F is applied to the edge of two hoops as the drawing below shows The hoops have the same mass whereas the radius of the larger hoop is twice the radius of the smaller one The entire...

Question

Answered step-by-step

user submitted image, transcription text available below
The same force $F$ is applied to the edge of two hoops as the drawing below shows. The hoops have the same mass, whereas the radius of the larger hoop is twice the radius of the smaller one. The entire mass of each hoop is concentrated at its rim, so the moment of inertia is equal to $\mathrm{Mr}^{2}$, where $M$ is the mass of the hoop, and $r$ is the radius. Which hoop has the greater angular acceleration, and how many times as great is it compared to the other hoop? The larger hoop; four times as great. The smaller hoop; four times as great. SUPPORT The larger hoop; two times as great.
· · ·
Report

🤔 Not the exact question you’re looking for?

Go ask your question

1 Solution

Best
Best
Newest
This problem has been solved! You'll receive a detailed solution to help you
master the concepts.

Solution 1

To determine which hoop has the greater angular acceleration and how many times greater it is compared to the other hoop, we need to analyze the situation using the concepts of torque, moment of inertia, and angular acceleration. #### Solution By Steps ***Step 1: Identify the Moment of Inertia for Each Hoop*** The moment of inertia (I) for a hoop is given by the formula: I = M * r^2 For the smaller hoop (radius = R): I_small = M * R^2 For the larger hoop (radius = 2R): I_large = M * (2R)^2 = M * 4R^2 ***Step 2: Calculate the Torque for Each Hoop*** Torque (τ) is given by the formula: τ = F * r For the smaller hoop: τ_small = F * R For the larger hoop: τ_large = F * (2R) = 2F * R ***Step 3: Relate Torque to Angular Acceleration*** Using the relationship between torque, moment of inertia, and angular acceleration (α): τ = I * α For the smaller hoop: F * R = (M * R^2) * α_small => α_small = (F * R) / (M * R^2) = F / (M * R) For the larger hoop: 2F * R = (M * 4R^2) * α_large => α_large = (2F * R) / (M * 4R^2) = (2F) / (M * 4R) = F / (2M * R) ***Step 4: Compare Angular Accelerations*** Now we compare α_small and α_large: α_small = F / (M * R) α_large = F / (2M * R) To find how many times greater α_small is than α_large: α_small / α_large = (F / (M * R)) / (F / (2M * R)) = 2 Thus, the smaller hoop has an angular acceleration that is 2 times greater than that of the larger hoop. #### Final Answer The smaller hoop; two times as great.
Answered by StudyX AI with Basic Model
Ask Tutor
Copy

📢 Boost your learning 10x faster with our browser extension! Effortlessly integrate it into any LMS like Canvas, Blackboard, Moodle and Pearson. Install now and revolutionize your study experience!

Ask a new question for Free

By text

By image

Drop file here or Click Here to upload
Ctrl
+
to upload
Try the StudyX app
Scan questions and equations with no typing!